Euler's Eyes
2026-10-24
Exercises from chapter VII — Equations
For each equation, first ask: what fundamental equation does it reduce to? After writing the families of solutions, check that none of them cancels out a denominator in the original equation.
Equation sin α = sin β
Solve the equations: a) $\operatorname{sen} x = \operatorname{sen}\dfrac{\pi}{5}$; b) $\operatorname{cossec} x = \operatorname{cossec}\dfrac{2\pi}{3}$; c) $\operatorname{sen} x = 0$; d) $\operatorname{sen} x = \dfrac{1}{2}$; e) $\operatorname{sen} x = -\dfrac{\sqrt{2}}{2}$; f) $\operatorname{sen} x = \dfrac{\sqrt{3}}{2}$; g) $\operatorname{sen} x = 1$; h) $\operatorname{sen} x = -1$.
See the resolution
a) $\operatorname{sen} x = \operatorname{sen}\dfrac{\pi}{5} \Rightarrow x = \dfrac{\pi}{5} + 2k\pi$ or $x = \pi - \dfrac{\pi}{5} + 2k\pi$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{5} + 2k\pi \text{ ou } x = \dfrac{4\pi}{5} + 2k\pi\right\}$.
b) $\operatorname{cossec} x = \operatorname{cossec}\dfrac{2\pi}{3} \Rightarrow \dfrac{1}{\operatorname{sen} x} = \dfrac{1}{\operatorname{sen}\frac{2\pi}{3}} \Rightarrow \operatorname{sen} x = \operatorname{sen}\dfrac{2\pi}{3}$, then $x = \dfrac{2\pi}{3} + 2k\pi$ or $x = \pi - \dfrac{2\pi}{3} + 2k\pi$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{2\pi}{3} + 2k\pi \text{ ou } x = \dfrac{\pi}{3} + 2k\pi\right\}$.
c) $\operatorname{sen} x = 0 = \operatorname{sen} 0 \Rightarrow x = 2k\pi$ or $x = \pi + 2k\pi = (2k + 1)\pi$. $\;S = \{x \in \mathbb{R} \mid x = k\pi\}$.
d) $\operatorname{sen} x = \dfrac{1}{2} = \operatorname{sen}\dfrac{\pi}{6}$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{6} + 2k\pi \text{ ou } x = \dfrac{5\pi}{6} + 2k\pi\right\}$.
e) $\operatorname{sen} x = -\dfrac{\sqrt{2}}{2} = \operatorname{sen}\dfrac{5\pi}{4}$, then $x = \dfrac{5\pi}{4} + 2k\pi$ or $x = \pi - \dfrac{5\pi}{4} + 2k\pi$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{5\pi}{4} + 2k\pi \text{ ou } x = -\dfrac{\pi}{4} + 2k\pi\right\}$.
f) $\operatorname{sen} x = \dfrac{\sqrt{3}}{2} = \operatorname{sen}\dfrac{\pi}{3}$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{3} + 2k\pi \text{ ou } x = \dfrac{2\pi}{3} + 2k\pi\right\}$.
g) $\operatorname{sen} x = 1 = \operatorname{sen}\dfrac{\pi}{2}$ (the two families coincide). $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{2} + 2k\pi\right\}$.
h) $\operatorname{sen} x = -1 = \operatorname{sen}\dfrac{3\pi}{2}$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{3\pi}{2} + 2k\pi\right\}$.
Solve the equations: a) $\operatorname{sen}^2 x = \dfrac{1}{4}$; b) $\operatorname{sen}^2 x - \operatorname{sen} x = 0$; c) $2\operatorname{sen}^2 x - 3\operatorname{sen} x + 1 = 0$; d) $2\cos^2 x = 1 - \operatorname{sen} x$.
See the resolution
a) $\operatorname{sen} x = \pm\dfrac{1}{2}$, then $S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{6} + 2k\pi \text{ ou } x = \dfrac{5\pi}{6} + 2k\pi \text{ ou } x = \dfrac{7\pi}{6} + 2k\pi \text{ ou } x = -\dfrac{\pi}{6} + 2k\pi\right\}$.
b) $\operatorname{sen} x(\operatorname{sen} x - 1) = 0 \Rightarrow \operatorname{sen} x = 0$ or $\operatorname{sen} x = 1$, then $S = \left\{x \in \mathbb{R} \;\middle|\; x = k\pi \text{ ou } x = \dfrac{\pi}{2} + 2k\pi\right\}$.
c) $\operatorname{sen} x = \dfrac{3 \pm \sqrt{9 - 8}}{4} = \dfrac{3 \pm 1}{4} \Rightarrow \operatorname{sen} x = 1$ or $\operatorname{sen} x = \dfrac{1}{2}$, then $S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{2} + 2k\pi \text{ ou } x = \dfrac{\pi}{6} + 2k\pi \text{ ou } x = \dfrac{5\pi}{6} + 2k\pi\right\}$.
d) $2(1 - \operatorname{sen}^2 x) = 1 - \operatorname{sen} x \Rightarrow 2\operatorname{sen}^2 x - \operatorname{sen} x - 1 = 0 \Rightarrow \operatorname{sen} x = \dfrac{1 \pm \sqrt{1 + 8}}{4} = \dfrac{1 \pm 3}{4} = 1$ or $-\dfrac{1}{2}$. We fall back on fundamental equations: $\operatorname{sen} x = 1 \Rightarrow x = \dfrac{\pi}{2} + 2k\pi$; $\;\operatorname{sen} x = -\dfrac{1}{2} \Rightarrow x = -\dfrac{\pi}{6} + 2k\pi$ or $x = \dfrac{7\pi}{6} + 2k\pi$.
$S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{2} + 2k\pi \text{ ou } x = -\dfrac{\pi}{6} + 2k\pi \text{ ou } x = \dfrac{7\pi}{6} + 2k\pi\right\}$ ($k \in \mathbb{Z}$).
Solve the equations: a) $\operatorname{sen} x = \operatorname{sen}\dfrac{\pi}{7}$; b) $\operatorname{cossec} x = 2$; c) $\operatorname{sen} x = -\dfrac{\sqrt{3}}{2}$; d) $2\operatorname{sen}^2 x = 1$; e) $\operatorname{sen}^2 x = 1$; f) $2\operatorname{sen}^2 x + \operatorname{sen} x - 1 = 0$; g) $2\operatorname{sen} x - \operatorname{cossec} x = 1$; h) $3\operatorname{tg} x = 2\cos x$; i) $\operatorname{sen} x + \cos 2x = 1$; j) $\cos^2 x = 1 - \operatorname{sen} x$.
Ver resposta
In all, $k \in \mathbb{Z}$. a) $x = \dfrac{\pi}{7} + 2k\pi$ or $x = \dfrac{6\pi}{7} + 2k\pi$ · b) $x = \dfrac{\pi}{6} + 2k\pi$ or $x = \dfrac{5\pi}{6} + 2k\pi$ · c) $x = \dfrac{4\pi}{3} + 2k\pi$ or $x = -\dfrac{\pi}{3} + 2k\pi$ · d) $x = \pm\dfrac{\pi}{4} + 2k\pi$, $x = \dfrac{3\pi}{4} + 2k\pi$ or $x = \dfrac{5\pi}{4} + 2k\pi$ · e) $x = \dfrac{\pi}{2} + 2k\pi$ or $x = \dfrac{3\pi}{2} + 2k\pi$ · f) $x = \dfrac{3\pi}{2} + 2k\pi$, $x = \dfrac{\pi}{6} + 2k\pi$ or $x = \dfrac{5\pi}{6} + 2k\pi$ · g) $x = \dfrac{\pi}{2} + 2k\pi$, $x = \dfrac{7\pi}{6} + 2k\pi$ or $x = -\dfrac{\pi}{6} + 2k\pi$ · h) $x = \dfrac{\pi}{6} + 2k\pi$ or $x = \dfrac{5\pi}{6} + 2k\pi$ · i) $x = k\pi$, $x = \dfrac{\pi}{6} + 2k\pi$ or $x = \dfrac{5\pi}{6} + 2k\pi$ · j) $x = k\pi$ or $x = \dfrac{\pi}{2} + 2k\pi$
(FEFAAP-77) Determine the values of $x$ that satisfy the equation $4\operatorname{sen}^4 x - 11\operatorname{sen}^2 x + 6 = 0$.
Ver resposta
$S = \left\{x \in \mathbb{R} \;\middle|\; x = \pm\dfrac{\pi}{3} + k\pi\right\}$
(FEI-76) Solve equation $2\operatorname{sen} x\,|\operatorname{sen} x| + 3\operatorname{sen} x = 2$.
Ver resposta
$S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{6} + 2k\pi \text{ ou } x = \dfrac{5\pi}{6} + 2k\pi\right\}$
Solve the equations: a) $\operatorname{sen} 2x = \dfrac{1}{2}$; b) $\operatorname{sen} 3x = \dfrac{\sqrt{2}}{2}$; c) $\operatorname{sen}\left(x - \dfrac{\pi}{3}\right) = \dfrac{\sqrt{3}}{2}$; d) $\operatorname{sen} 2x = \operatorname{sen} x$.
See the resolution
a) $\operatorname{sen} 2x = \operatorname{sen}\dfrac{\pi}{6} \Rightarrow 2x = \dfrac{\pi}{6} + 2k\pi$ or $2x = \pi - \dfrac{\pi}{6} + 2k\pi$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{12} + k\pi \text{ ou } x = \dfrac{5\pi}{12} + k\pi\right\}$.
b) $\operatorname{sen} 3x = \operatorname{sen}\dfrac{\pi}{4} \Rightarrow 3x = \dfrac{\pi}{4} + 2k\pi$ or $3x = \pi - \dfrac{\pi}{4} + 2k\pi$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{12} + \dfrac{2k\pi}{3} \text{ ou } x = \dfrac{\pi}{4} + \dfrac{2k\pi}{3}\right\}$.
c) $\operatorname{sen}\left(x - \dfrac{\pi}{3}\right) = \operatorname{sen}\dfrac{\pi}{3} \Rightarrow x - \dfrac{\pi}{3} = \dfrac{\pi}{3} + 2k\pi$ or $x - \dfrac{\pi}{3} = \pi - \dfrac{\pi}{3} + 2k\pi$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{2\pi}{3} + 2k\pi \text{ ou } x = \pi + 2k\pi\right\}$.
d) $\operatorname{sen} 2x = \operatorname{sen} x \Rightarrow 2x = x + 2k\pi$ or $2x = \pi - x + 2k\pi$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = 2k\pi \text{ ou } x = \dfrac{\pi}{3} + \dfrac{2k\pi}{3}\right\}$.
Determine $x \in \mathbb{R}$ such that: a) $\operatorname{sen} 5x = \operatorname{sen} 3x$; b) $\operatorname{sen} 3x = \operatorname{sen} 2x$.
Ver resposta
a) $\left\{x \in \mathbb{R} \;\middle|\; x = k\pi \text{ ou } x = \dfrac{\pi}{8} + \dfrac{k\pi}{4}\right\}$ · b) $\left\{x \in \mathbb{R} \;\middle|\; x = 2k\pi \text{ ou } x = \dfrac{\pi}{5} + \dfrac{2k\pi}{5}\right\}$
Determine the internal angles of a triangle $ABC$ knowing that they are in arithmetic progression and that the sine of the sum of the smallest angle and the average angle is $\dfrac{\sqrt{3}}{2}$.
Ver resposta
$A = B = C = \dfrac{\pi}{3}$ (triangle is equilateral)
(MAPOFEI-76) Solve system $\begin{cases} \operatorname{sen}(x + y) = 0 \\ x - y = \pi \end{cases}$
Ver resposta
$x = \dfrac{\pi}{2} + \dfrac{k\pi}{2}$, $y = -\dfrac{\pi}{2} + \dfrac{k\pi}{2}$ ($k \in \mathbb{Z}$)
Equation cos α = cos β
Solve the equations: a) $\cos x = \cos\dfrac{\pi}{5}$; b) $\sec x = \sec\dfrac{2\pi}{3}$; c) $\cos x = 0$; d) $\cos x = 1$; e) $\cos x = -1$; f) $\cos x = \dfrac{1}{2}$; g) $\cos x = \dfrac{\sqrt{2}}{2}$; h) $\cos x = -\dfrac{\sqrt{3}}{2}$.
See the resolution
a) $\cos x = \cos\dfrac{\pi}{5} \Rightarrow x = \pm\dfrac{\pi}{5} + 2k\pi$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \pm\dfrac{\pi}{5} + 2k\pi\right\}$.
b) $\dfrac{1}{\cos x} = \dfrac{1}{\cos\frac{2\pi}{3}} \Rightarrow \cos x = \cos\dfrac{2\pi}{3}$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \pm\dfrac{2\pi}{3} + 2k\pi\right\}$.
c) $\cos x = 0 = \cos\dfrac{\pi}{2}$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \pm\dfrac{\pi}{2} + 2k\pi\right\} = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{2} + k\pi\right\}$.
d) $\cos x = 1 = \cos 0$. $\;S = \{x \in \mathbb{R} \mid x = 2k\pi\}$. $\qquad$ e) $\cos x = -1 = \cos\pi$. $\;S = \{x \in \mathbb{R} \mid x = \pi + 2k\pi\}$.
f) $\cos x = \dfrac{1}{2} = \cos\dfrac{\pi}{3}$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \pm\dfrac{\pi}{3} + 2k\pi\right\}$. $\qquad$ g) $\cos x = \cos\dfrac{\pi}{4}$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \pm\dfrac{\pi}{4} + 2k\pi\right\}$.
h) $\cos x = -\dfrac{\sqrt{3}}{2} = \cos\dfrac{5\pi}{6}$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \pm\dfrac{5\pi}{6} + 2k\pi\right\}$.
Solve the equations: a) $4\cos^2 x = 3$; b) $\cos^2 x + \cos x = 0$; c) $\operatorname{sen}^2 x = 1 + \cos x$; d) $\cos 2x + 3\cos x + 2 = 0$.
See the resolution
a) $\cos^2 x = \dfrac{3}{4} \Rightarrow \cos x = \dfrac{\sqrt{3}}{2}$ or $\cos x = -\dfrac{\sqrt{3}}{2}$, then $S = \left\{x \in \mathbb{R} \;\middle|\; x = \pm\dfrac{\pi}{6} + 2k\pi \text{ ou } x = \pm\dfrac{5\pi}{6} + 2k\pi\right\}$.
b) $\cos x(\cos x + 1) = 0 \Rightarrow \cos x = 0$ or $\cos x = -1$, then $S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{2} + k\pi \text{ ou } x = \pi + 2k\pi\right\}$.
c) $1 - \cos^2 x = 1 + \cos x \Rightarrow \cos^2 x + \cos x = 0$, and we return to the previous item.
d) $(2\cos^2 x - 1) + 3\cos x + 2 = 0 \Rightarrow 2\cos^2 x + 3\cos x + 1 = 0 \Rightarrow \cos x = \dfrac{-3 \pm \sqrt{9 - 8}}{4} = \dfrac{-3 \pm 1}{4}$, i.e. $\cos x = -1$ or $\cos x = -\dfrac{1}{2}$. Then $S = \left\{x \in \mathbb{R} \;\middle|\; x = \pi + 2k\pi \text{ ou } x = \pm\dfrac{2\pi}{3} + 2k\pi\right\}$.
Solve the equations: a) $\cos x = -\dfrac{1}{2}$; b) $\cos x = -\dfrac{\sqrt{2}}{2}$; c) $\cos x = \dfrac{\sqrt{3}}{2}$; d) $\sec x = 2$; e) $2\cos^2 x = \cos x$; f) $4\cos x + 3\sec x = 8$; g) $2 - 2\cos x = \operatorname{sen} x \operatorname{tg} x$; h) $2\operatorname{sen}^2 x + 6\cos x = 5 + \cos 2x$; i) $1 + 3\operatorname{tg}^2 x = 5\sec x$; j) $\left(4 - \dfrac{3}{\operatorname{sen}^2 x}\right)\left(4 - \dfrac{1}{\cos^2 x}\right) = 0$.
Ver resposta
a) $x = \pm\dfrac{2\pi}{3} + 2k\pi$ · b) $x = \pm\dfrac{3\pi}{4} + 2k\pi$ · c) $x = \pm\dfrac{\pi}{6} + 2k\pi$ · d) $x = \pm\dfrac{\pi}{3} + 2k\pi$ · e) $x = \pm\dfrac{\pi}{3} + 2k\pi$ or $x = \dfrac{\pi}{2} + k\pi$ · f) $x = \pm\dfrac{\pi}{3} + 2k\pi$ · g) $x = 2k\pi$ · h) $x = 2k\pi$ or $x = \pm\dfrac{\pi}{3} + 2k\pi$ · i) $x = \pm\dfrac{\pi}{3} + 2k\pi$ · j) $x = \pm\dfrac{\pi}{3} + 2k\pi$ or $x = \pm\dfrac{2\pi}{3} + 2k\pi$
Solve the equations: a) $\cos 2x = \dfrac{\sqrt{3}}{2}$; b) $\cos 2x = \cos x$; c) $\cos\left(x + \dfrac{\pi}{6}\right) = 0$; d) $\cos\left(x - \dfrac{\pi}{4}\right) = 1$.
See the resolution
a) $\cos 2x = \cos\dfrac{\pi}{6} \Rightarrow 2x = \pm\dfrac{\pi}{6} + 2k\pi$, then $S = \left\{x \in \mathbb{R} \;\middle|\; x = \pm\dfrac{\pi}{12} + k\pi\right\}$.
b) $\cos 2x = \cos x \Rightarrow 2x = x + 2k\pi$ or $2x = -x + 2k\pi$, then $S = \left\{x \in \mathbb{R} \;\middle|\; x = 2k\pi \text{ ou } x = \dfrac{2k\pi}{3}\right\}$ (the first family is contained in the second).
c) $\cos\left(x + \dfrac{\pi}{6}\right) = 0 = \cos\dfrac{\pi}{2} \Rightarrow x + \dfrac{\pi}{6} = \pm\dfrac{\pi}{2} + 2k\pi$, then $S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{3} + 2k\pi \text{ ou } x = -\dfrac{2\pi}{3} + 2k\pi\right\}$.
d) $\cos\left(x - \dfrac{\pi}{4}\right) = 1 = \cos 0 \Rightarrow x - \dfrac{\pi}{4} = 2k\pi$, then $S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{4} + 2k\pi\right\}$.
Solve the equations: a) $\cos 3x - \cos x = 0$; b) $\cos 5x = \cos\left(x + \dfrac{\pi}{3}\right)$.
Ver resposta
a) $S = \left\{x \in \mathbb{R} \;\middle|\; x = k\pi \text{ ou } x = \dfrac{k\pi}{2}\right\}$ · b) $S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{12} + \dfrac{k\pi}{2} \text{ ou } x = -\dfrac{\pi}{18} + \dfrac{k\pi}{3}\right\}$
Determine the interior angles of a triangle $ABC$ knowing that $\cos(A + B) = \dfrac{1}{2}$ and $\operatorname{sen}(B + C) = \dfrac{1}{2}$.
Ver resposta
$A = \dfrac{\pi}{6}$, $B = \dfrac{\pi}{6}$, $C = \dfrac{2\pi}{3}$
(MAUÁ-77) Given the equation $(\operatorname{sen} x + \cos y)(\sec x + \operatorname{cossec} y) = 4$: a) solve it if $x = y$; b) solve it if $\operatorname{sen} x = \cos y$.
Ver resposta
a) $x = y = \dfrac{\pi}{4} + k\pi$ · b) $x = \dfrac{\pi}{4} + k\pi$ e $y + x = \dfrac{\pi}{2} + 2k\pi$
Equation tg α = tg β
Solve the equations: a) $\operatorname{tg} x = 1$; b) $\operatorname{cotg} x = \sqrt{3}$; c) $\operatorname{tg} x = -\sqrt{3}$; d) $\operatorname{tg} x = 0$; e) $\operatorname{tg} 2x = \sqrt{3}$; f) $\operatorname{tg} 2x = \operatorname{tg} x$; g) $\operatorname{tg} 3x = 1$; h) $\operatorname{tg} 5x = \operatorname{tg} 3x$.
See the resolution
a) $\operatorname{tg} x = 1 = \operatorname{tg}\dfrac{\pi}{4}$, then $S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{4} + k\pi\right\}$.
b) $\operatorname{cotg} x = \sqrt{3} \Rightarrow \operatorname{tg} x = \dfrac{1}{\sqrt{3}} = \operatorname{tg}\dfrac{\pi}{6}$, then $S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{6} + k\pi\right\}$.
c) $\operatorname{tg} x = -\sqrt{3} = \operatorname{tg}\dfrac{2\pi}{3}$, then $S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{2\pi}{3} + k\pi\right\}$.
d) $\operatorname{tg} x = 0 = \operatorname{tg} 0$, then $S = \{x \in \mathbb{R} \mid x = k\pi\}$.
e) $\operatorname{tg} 2x = \sqrt{3} = \operatorname{tg}\dfrac{\pi}{3} \Rightarrow 2x = \dfrac{\pi}{3} + k\pi$, then $S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{6} + \dfrac{k\pi}{2}\right\}$.
f) $\operatorname{tg} 2x = \operatorname{tg} x \Rightarrow 2x = x + k\pi$, then $S = \{x \in \mathbb{R} \mid x = k\pi\}$.
g) $\operatorname{tg} 3x = 1 = \operatorname{tg}\dfrac{\pi}{4} \Rightarrow 3x = \dfrac{\pi}{4} + k\pi$, then $S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{12} + \dfrac{k\pi}{3}\right\}$.
h) $\operatorname{tg} 5x = \operatorname{tg} 3x \Rightarrow 5x = 3x + k\pi \Rightarrow x = \dfrac{k\pi}{2}$. If $k$ is odd, $\operatorname{tg} 5x$ and $\operatorname{tg} 3x$ do not exist; therefore $S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{k\pi}{2},\ k \text{ par}\right\} = \{x \in \mathbb{R} \mid x = k\pi\}$.
Solve the equations: a) $\operatorname{sen} x - \sqrt{3}\cos x = 0$; b) $\operatorname{sen}^2 x = \cos^2 x$; c) $\operatorname{tg} x + \operatorname{cotg} x = 2$; d) $\sec^2 x = 1 + \operatorname{tg} x$.
See the resolution
a) $\operatorname{sen} x = \sqrt{3}\cos x \Rightarrow \dfrac{\operatorname{sen} x}{\cos x} = \sqrt{3} \Rightarrow \operatorname{tg} x = \sqrt{3}$ ($\cos x = 0$ is not useful, as it would also be $\operatorname{sen} x = 0$). $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{3} + k\pi\right\}$.
b) $\dfrac{\operatorname{sen}^2 x}{\cos^2 x} = 1 \Rightarrow \operatorname{tg}^2 x = 1$, then $\operatorname{tg} x = 1$ or $\operatorname{tg} x = -1$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{4} + k\pi \text{ ou } x = \dfrac{3\pi}{4} + k\pi\right\}$.
c) $\operatorname{tg} x + \dfrac{1}{\operatorname{tg} x} = 2 \Rightarrow \operatorname{tg}^2 x - 2\operatorname{tg} x + 1 = 0 \Rightarrow \operatorname{tg} x = \dfrac{2 \pm \sqrt{4 - 4}}{2} = 1$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{4} + k\pi\right\}$.
d) $1 + \operatorname{tg}^2 x = 1 + \operatorname{tg} x \Rightarrow \operatorname{tg} x(\operatorname{tg} x - 1) = 0$, then $\operatorname{tg} x = 0$ or $\operatorname{tg} x = 1$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = k\pi \text{ ou } x = \dfrac{\pi}{4} + k\pi\right\}$.
Solve the equations: a) $\operatorname{tg} x = \operatorname{tg}\dfrac{\pi}{5}$; b) $\operatorname{cotg} x = \operatorname{cotg}\dfrac{5\pi}{6}$; c) $3\operatorname{tg} x = \sqrt{3}$; d) $\operatorname{cotg} x = 0$; e) $\operatorname{cotg} x = -1$; f) $\operatorname{tg} 3x - \operatorname{tg} 2x = 0$; g) $\operatorname{tg} 2x = \operatorname{tg}\left(x + \dfrac{\pi}{4}\right)$; h) $\operatorname{tg} 4x = 1$; i) $\operatorname{cotg} 2x = \operatorname{cotg}\left(x + \dfrac{\pi}{4}\right)$; j) $\operatorname{tg}^2 2x = 3$.
Ver resposta
a) $x = \dfrac{\pi}{5} + k\pi$ · b) $x = \dfrac{5\pi}{6} + k\pi$ · c) $x = \dfrac{\pi}{6} + k\pi$ · d) $x = \dfrac{\pi}{2} + k\pi$ · e) $x = \dfrac{3\pi}{4} + k\pi$ · f) $x = k\pi$ · g) $S = \varnothing$ · h) $x = \dfrac{\pi}{16} + \dfrac{k\pi}{4}$ · i) $x = \dfrac{\pi}{4} + k\pi$ · j) $x = \dfrac{\pi}{6} + \dfrac{k\pi}{2}$ or $x = \dfrac{\pi}{3} + \dfrac{k\pi}{2}$
Solve the equations: a) $\sec^2 x = 2\operatorname{tg} x$; b) $\dfrac{1}{\operatorname{sen}^2 x} = 1 - \dfrac{\cos x}{\operatorname{sen} x}$; c) $\operatorname{sen} 2x\cos\left(x + \dfrac{\pi}{4}\right) = \cos 2x\operatorname{sen}\left(x + \dfrac{\pi}{4}\right)$; d) $(1 - \operatorname{tg} x)(1 + \operatorname{sen} 2x) = 1 + \operatorname{tg} x$.
Ver resposta
a) $x = \dfrac{\pi}{4} + k\pi$ · b) $x = \dfrac{\pi}{2} + k\pi$ or $x = \dfrac{3\pi}{4} + k\pi$ · c) $x = \dfrac{\pi}{4} + k\pi$ · d) $x = \dfrac{3\pi}{4} + k\pi$ or $x = k\pi$
(MAPOFEI-75) Solve the equation $\operatorname{cotg} x - \operatorname{sen} 2x = 0$.
Ver resposta
$S = \left\{x \in \mathbb{R} \;\middle|\; x = \pm\dfrac{\pi}{4} + k\pi \text{ ou } x = \dfrac{\pi}{2} + k\pi\right\}$. Indeed, $\dfrac{\cos x}{\operatorname{sen} x} - 2\operatorname{sen} x\cos x = \dfrac{\cos x(1 - 2\operatorname{sen}^2 x)}{\operatorname{sen} x} = \dfrac{\cos x\cos 2x}{\operatorname{sen} x} = 0$. (The printed edition template only has $\pm\frac{\pi}{4} + k\pi$, omitting the roots with $\cos x = 0$.)
(FEI-77) For what values of $p$ does the equation $\operatorname{tg} px = \operatorname{cotg} px$ have $x = \dfrac{\pi}{2}$ as its root?
Ver resposta
$p = \dfrac{1}{2} + k$, $k \in \mathbb{Z}$
Solutions in a range
Determine $x \in [0, 2\pi]$ such that $2\operatorname{sen} x = 1$.
See the resolution
The general solution of equation $\operatorname{sen} x = \dfrac{1}{2}$ is $x = \dfrac{\pi}{6} + 2k\pi$ or $x = \dfrac{5\pi}{6} + 2k\pi$. To have $0 \le x \le 2\pi$, we must assign $k$ the value 0; then $S = \left\{\dfrac{\pi}{6}, \dfrac{5\pi}{6}\right\}$.
What are the arcs of the closed interval $[0, 2\pi]$ such that the sine of their double is $\dfrac{\sqrt{3}}{2}$?
See the resolution
Let us call the sought arcs $x$:
$$\operatorname{sen} 2x = \frac{\sqrt{3}}{2} = \operatorname{sen}\frac{\pi}{3} \;\Longrightarrow\; 2x = \frac{\pi}{3} + 2k\pi \;\text{ ou }\; 2x = \frac{2\pi}{3} + 2k\pi \;\Longrightarrow\; x = \frac{\pi}{6} + k\pi \;(1) \;\text{ ou }\; x = \frac{\pi}{3} + k\pi \;(2)$$(the so-called general solution). We only want the arcs of $[0, 2\pi]$; we assign values to $k$: in (1), $k = 0 \Rightarrow x = \frac{\pi}{6}$ and $k = 1 \Rightarrow x = \frac{7\pi}{6}$; in (2), $k = 0 \Rightarrow x = \frac{\pi}{3}$ and $k = 1 \Rightarrow x = \frac{4\pi}{3}$. Any other value of $k$ would take $x$ out of $[0, 2\pi]$.
$$S = \left\{\frac{\pi}{6}, \frac{\pi}{3}, \frac{7\pi}{6}, \frac{4\pi}{3}\right\}$$Determine $x$ such that $0 < x < \pi$ and $\operatorname{sen} 3x = \dfrac{1}{2}$.
Ver resposta
$S = \left\{\dfrac{\pi}{18}, \dfrac{5\pi}{18}, \dfrac{13\pi}{18}, \dfrac{17\pi}{18}\right\}$
What are the arcs of the closed interval $[0, 2\pi]$ such that their sine is equal to the sine of their double?
See the resolution
In (1): $k = 0 \Rightarrow x = 0$; $k = 1 \Rightarrow x = 2\pi$. In (2): $k = 0 \Rightarrow x = \frac{\pi}{3}$; $k = 1 \Rightarrow x = \pi$; $k = 2 \Rightarrow x = \frac{5\pi}{3}$.
$$S = \left\{0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}, 2\pi\right\}$$Obtain the solutions of the equation $\operatorname{sen} 3x = \operatorname{sen} 2x$ that belong to the interval $[0, \pi]$.
Ver resposta
$S = \left\{0, \dfrac{\pi}{5}, \dfrac{3\pi}{5}, \pi\right\}$
Determine $x$ such that $0 < x < 2\pi$ and $\cos 2x = \dfrac{1}{2}$.
See the resolution
From (1): $k = 0 \Rightarrow \frac{\pi}{6}$, $k = 1 \Rightarrow \frac{7\pi}{6}$. From (2): $k = 1 \Rightarrow \frac{5\pi}{6}$, $k = 2 \Rightarrow \frac{11\pi}{6}$. Then $S = \left\{\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}\right\}$.
Get $x$ such that $\cos 3x = \cos 2x$ and $0 \le x \le \pi$.
Ver resposta
$S = \left\{0, \dfrac{2\pi}{5}, \dfrac{4\pi}{5}\right\}$
(EESCUSP-69) Find the solutions from $4\operatorname{sen}^3 x - \operatorname{sen} x = 0$ to $0 \le x \le 2\pi$.
Ver resposta
$S = \left\{0, \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \pi, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}, 2\pi\right\}$
Determine $x$ such that $0 \le x \le \pi$ and $\operatorname{tg} 6x = \operatorname{tg} 2x$.
See the resolution
$\operatorname{tg} 6x = \operatorname{tg} 2x \Rightarrow 6x = 2x + k\pi \Rightarrow x = \dfrac{k\pi}{4}$. Doing $k = 0, 1, 2, 3, 4$, we obtain $x = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi$. Excluding $\frac{\pi}{4}$ and $\frac{3\pi}{4}$, for which the tangents of $6x$ and $2x$ do not exist:
$$S = \left\{0, \frac{\pi}{2}, \pi\right\}.$$(MAPOFEI-74) Calculate $x$ in the range $0 \le x \le 2\pi$ such that $\operatorname{tg} x + \operatorname{cotg} x = 2$.
Ver resposta
$S = \left\{\dfrac{\pi}{4}, \dfrac{5\pi}{4}\right\}$
Being $0 \le x \le \pi$, solve for $\sqrt{\operatorname{sen}^2 x} - \sqrt{\cos^2 x} = 0$.
Ver resposta
$S = \left\{\dfrac{\pi}{4}, \dfrac{3\pi}{4}\right\}$
(Itajubá-69) Solve the equation $\operatorname{sen} x + \operatorname{sen} y = 1$ knowing that $x + y = \dfrac{\pi}{3}$.
Ver resposta
$x = \dfrac{\pi}{6} + 2k\pi$ e $y = \dfrac{\pi}{6} - 2k\pi$
Classical equations
Solve the equation $\sqrt{3}\cos x + \operatorname{sen} x = 1$.
View resolution (three methods)
Method 1. Doing $\operatorname{sen} x = u$ and $\cos x = v$: $\begin{cases} u + \sqrt{3}\,v = 1 & (1) \\ u^2 + v^2 = 1 & (2) \end{cases}$. From (1), $u = 1 - \sqrt{3}\,v$, which substituted into (2) gives $(1 - \sqrt{3}\,v)^2 + v^2 = 1 \Rightarrow 4v^2 - 2\sqrt{3}\,v = 0$. Then $v = 0$ (and $u = 1$) or $v = \frac{\sqrt{3}}{2}$ (and $u = 1 - \frac{3}{2} = -\frac{1}{2}$). There are two possibilities: $\cos x = 0$ and $\operatorname{sen} x = 1$, that is, $x = \frac{\pi}{2} + 2k\pi$; or $\cos x = \frac{\sqrt{3}}{2}$ and $\operatorname{sen} x = -\frac{1}{2}$, that is, $x = -\frac{\pi}{6} + 2k\pi$.
Method 2. $\operatorname{sen} x + \sqrt{3}\cos x = 1 \Rightarrow \operatorname{sen} x + \operatorname{tg}\frac{\pi}{3}\cos x = 1 \Rightarrow \operatorname{sen} x + \dfrac{\operatorname{sen}\frac{\pi}{3}}{\cos\frac{\pi}{3}}\cos x = 1 \Rightarrow \operatorname{sen} x\cos\frac{\pi}{3} + \operatorname{sen}\frac{\pi}{3}\cos x = \cos\frac{\pi}{3}$
$\Rightarrow \operatorname{sen}\left(x + \frac{\pi}{3}\right) = \frac{1}{2} \Rightarrow x + \frac{\pi}{3} = \frac{\pi}{6} + 2k\pi$ or $x + \frac{\pi}{3} = \frac{5\pi}{6} + 2k\pi \Rightarrow x = -\frac{\pi}{6} + 2k\pi$ or $x = \frac{\pi}{2} + 2k\pi$.
Method 3. With $t = \operatorname{tg}\frac{x}{2}$: $\dfrac{2t}{1 + t^2} + \sqrt{3} \cdot \dfrac{1 - t^2}{1 + t^2} = 1 \Rightarrow 2t + \sqrt{3} - \sqrt{3}\,t^2 = 1 + t^2 \Rightarrow (1 + \sqrt{3})t^2 - 2t + (1 - \sqrt{3}) = 0$. Then
$$t = \frac{2 \pm \sqrt{4 - 4(1 + \sqrt{3})(1 - \sqrt{3})}}{2(1 + \sqrt{3})} = \frac{2 \pm 2\sqrt{3}}{2(1 + \sqrt{3})} = 1 \;\text{ ou }\; -2 + \sqrt{3}.$$$t = \operatorname{tg}\frac{x}{2} = 1 \Rightarrow \frac{x}{2} = \frac{\pi}{4} + k\pi \Rightarrow x = \frac{\pi}{2} + 2k\pi$; $\;t = -2 + \sqrt{3} = \operatorname{tg}\left(-\frac{\pi}{12}\right) \Rightarrow x = -\frac{\pi}{6} + 2k\pi$.
Solve the equations: a) $\operatorname{sen} x + \cos x = -1$; b) $\sqrt{3}\operatorname{sen} x - \cos x = -\sqrt{3}$.
Ver resposta
a) $S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{3\pi}{2} + 2k\pi \text{ ou } x = \pi + 2k\pi\right\}$ · b) $S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{11\pi}{6} + 2k\pi \text{ ou } x = \dfrac{3\pi}{2} + 2k\pi\right\}$
Determine $x$ such that $0 \le x \le 2\pi$ and $\operatorname{sen} x + \cos x = 1$.
See the resolution
Making $\operatorname{sen} x = u$ and $\cos x = v$: $\begin{cases} u + v = 1 & (1) \\ u^2 + v^2 = 1 & (2) \end{cases}$. (1) in (2): $u^2 + (1 - u)^2 = 1 \Rightarrow 2u^2 - 2u = 0$. There are two possibilities: $u = 0$ and $v = 1$, or $u = 1$ and $v = 0$. Therefore $S = \left\{0, \dfrac{\pi}{2}, 2\pi\right\}$.
Obtain the solutions of the equations within the range $[0, 2\pi]$: a) $\operatorname{sen} 4x + \cos 4x = 1$; b) $|\operatorname{sen} x| + |\cos x| = 1$.
Ver resposta
a) $S = \left\{0, \dfrac{\pi}{8}, \dfrac{\pi}{2}, \dfrac{5\pi}{8}, \pi, \dfrac{9\pi}{8}, \dfrac{3\pi}{2}, \dfrac{13\pi}{8}, 2\pi\right\}$ · b) $S = \left\{0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2}, 2\pi\right\}$
(MACK-70) Solve the equation $\operatorname{sen} 2x = 1 - \cos 2x$ in the set of real numbers.
Ver resposta
$S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{4} + k\pi \text{ ou } x = k\pi\right\}$
Discuss the equation in $x$: $m\operatorname{sen} x + \cos x = m$.
See the resolution
Doing $\operatorname{sen} x = \dfrac{2t}{1 + t^2}$ and $\cos x = \dfrac{1 - t^2}{1 + t^2}$:
$$m \cdot \frac{2t}{1 + t^2} + \frac{1 - t^2}{1 + t^2} = m \;\Longrightarrow\; 2mt + 1 - t^2 = m + mt^2 \;\Longrightarrow\; (m + 1)t^2 - 2mt + (m - 1) = 0.$$This equation has a real solution if, and only if, $\Delta \ge 0$: $\Delta = 4m^2 - 4(m + 1)(m - 1) = 4 > 0$, which occurs for every real $m$. (When $m = -1$ the equation in $t$ is of the 1st degree, $2t - 2 = 0$, and there is still a solution; and $x = \pi + 2k\pi$, for which the substitution is not valid, is only a root if $m = -1$.) Therefore, the equation has a solution for todo $m \in \mathbb{R}$.
Discuss, according to $m$, the equations: a) $m\cos x - (m + 1)\operatorname{sen} x = m$; b) $\operatorname{sen} x + \cos x = m$.
Ver resposta
a) there is a solution for all $m \in \mathbb{R}$ · b) there is a solution if, and only if, $-\sqrt{2} \le m \le \sqrt{2}$
Solve the equations: a) $\operatorname{sen} 7x + \operatorname{sen} 5x = 0$; b) $\cos 6x + \cos 2x = 0$; c) $\operatorname{sen} 4x - \cos x = 0$; d) $\cos 3x + \operatorname{sen} 2x = 0$.
See the resolution
a) $\operatorname{sen} 7x + \operatorname{sen} 5x = 0 \Rightarrow 2\operatorname{sen} 6x\cos x = 0$. 1st possibility: $\operatorname{sen} 6x = 0 \Rightarrow 6x = k\pi \Rightarrow x = \frac{k\pi}{6}$. 2nd possibility: $\cos x = 0 \Rightarrow x = \frac{\pi}{2} + k\pi$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{k\pi}{6} \text{ ou } x = \dfrac{\pi}{2} + k\pi\right\}$.
b) $\cos 6x + \cos 2x = 0 \Rightarrow 2\cos 4x\cos 2x = 0$. $\cos 4x = 0 \Rightarrow x = \frac{\pi}{8} + \frac{k\pi}{4}$; $\cos 2x = 0 \Rightarrow x = \frac{\pi}{4} + \frac{k\pi}{2}$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{8} + \dfrac{k\pi}{4} \text{ ou } x = \dfrac{\pi}{4} + \dfrac{k\pi}{2}\right\}$.
c) $\operatorname{sen} 4x - \operatorname{sen}\left(\frac{\pi}{2} - x\right) = 0 \Rightarrow 2\operatorname{sen}\left(\frac{5x}{2} - \frac{\pi}{4}\right)\cos\left(\frac{3x}{2} + \frac{\pi}{4}\right) = 0$. 1ª: $\frac{5x}{2} - \frac{\pi}{4} = k\pi \Rightarrow x = \frac{\pi}{10} + \frac{2k\pi}{5}$. 2ª: $\frac{3x}{2} + \frac{\pi}{4} = \frac{\pi}{2} + k\pi \Rightarrow x = \frac{\pi}{6} + \frac{2k\pi}{3}$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{10} + \dfrac{2k\pi}{5} \text{ ou } x = \dfrac{\pi}{6} + \dfrac{2k\pi}{3}\right\}$.
d) $\cos 3x + \cos\left(\frac{\pi}{2} - 2x\right) = 0 \Rightarrow 2\cos\left(\frac{x}{2} + \frac{\pi}{4}\right)\cos\left(\frac{5x}{2} - \frac{\pi}{4}\right) = 0$. 1ª: $\frac{x}{2} + \frac{\pi}{4} = \frac{\pi}{2} + k\pi \Rightarrow x = \frac{\pi}{2} + 2k\pi$. 2ª: $\frac{5x}{2} - \frac{\pi}{4} = \frac{\pi}{2} + k\pi \Rightarrow x = \frac{3\pi}{10} + \frac{2k\pi}{5}$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{2} + 2k\pi \text{ ou } x = \dfrac{3\pi}{10} + \dfrac{2k\pi}{5}\right\}$.
Solve the equations: a) $\operatorname{sen} mx + \operatorname{sen} nx = 0$ ($m, n \in \mathbb{N}^*$); b) $\cos ax + \cos bx = 0$ ($a, b \in \mathbb{R}^*$); c) $\operatorname{sen} 2x = \cos\left(x + \dfrac{\pi}{4}\right)$.
Ver resposta
a) $x = \dfrac{2k\pi}{m + n}$ or $x = \dfrac{(2k + 1)\pi}{m - n}$ (the latter only if $m \ne n$) · b) $x = \dfrac{\pi}{a + b} + \dfrac{2k\pi}{a + b}$ or $x = -\dfrac{\pi}{a - b} + \dfrac{2k\pi}{a - b}$ (with $a + b \ne 0$ and $a - b \ne 0$, respectively) · c) $x = \dfrac{\pi}{12} + \dfrac{2k\pi}{3}$ or $x = \dfrac{3\pi}{4} + 2k\pi$
Solve the equations: a) $\operatorname{sen} x + \operatorname{sen} 3x + \operatorname{sen} 4x + \operatorname{sen} 6x = 0$; b) $\cos 3x + \cos 7x = \cos 5x$.
See the resolution
a) $(\operatorname{sen} 6x + \operatorname{sen} 4x) + (\operatorname{sen} 3x + \operatorname{sen} x) = 0 \Rightarrow 2\operatorname{sen} 5x\cos x + 2\operatorname{sen} 2x\cos x = 0 \Rightarrow \cos x(\operatorname{sen} 5x + \operatorname{sen} 2x) = 0 \Rightarrow 2\cos x\operatorname{sen}\dfrac{7x}{2}\cos\dfrac{3x}{2} = 0$.
1st possibility: $\cos x = 0 \Rightarrow x = \frac{\pi}{2} + k\pi$. 2nd: $\operatorname{sen}\frac{7x}{2} = 0 \Rightarrow x = \frac{2k\pi}{7}$. 3rd: $\cos\frac{3x}{2} = 0 \Rightarrow x = \frac{\pi}{3} + \frac{2k\pi}{3}$.
$S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{2} + k\pi,\ x = \dfrac{2k\pi}{7} \text{ ou } x = \dfrac{\pi}{3} + \dfrac{2k\pi}{3}\right\}$.
b) $(\cos 7x + \cos 3x) - \cos 5x = 0 \Rightarrow 2\cos 5x\cos 2x - \cos 5x = 0 \Rightarrow 2\cos 5x\left(\cos 2x - \frac{1}{2}\right) = 0$.
1ª: $\cos 5x = 0 \Rightarrow x = \frac{\pi}{10} + \frac{k\pi}{5}$. 2ª: $\cos 2x = \frac{1}{2} \Rightarrow x = \pm\frac{\pi}{6} + k\pi$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{10} + \dfrac{k\pi}{5} \text{ ou } x = \pm\dfrac{\pi}{6} + k\pi\right\}$.
Solve the equations: a) $\operatorname{sen} 5x + \operatorname{sen} x = 2\operatorname{sen} 3x$; b) $\cos x + \cos(2x + a) + \cos(3x + 2a) = 0$; c) $\operatorname{sen} 7x + \cos 3x = \cos 5x - \operatorname{sen} x$.
Ver resposta
a) $x = k\pi$ or $x = \dfrac{k\pi}{3}$ · b) $x = \pm\dfrac{\pi}{4} - \dfrac{a}{2} + k\pi$, $x = \dfrac{2\pi}{3} - a + 2k\pi$ or $x = \dfrac{4\pi}{3} - a + 2k\pi$ · c) $x = \dfrac{k\pi}{4}$ or $x = -\dfrac{\pi}{8} + \dfrac{k\pi}{2}$
Determine $x$ such that $0 \le x \le \pi$ and $\cos^2(x + a) + \cos^2(x - a) = 1$.
Ver resposta
$S = \left\{\dfrac{\pi}{4}, \dfrac{3\pi}{4}\right\}$ (assuming $\cos 2a \ne 0$; if $\cos 2a = 0$, all $x$ will do)
Determine $x$ such that $\operatorname{sen} 3x + \cos 2x - \operatorname{sen} x = 1$ and $0 \le x \le \pi$.
Ver resposta
$S = \left\{0, \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \pi\right\}$
(MAPOFEI-74) Determine the angle $x$, measured in radians, that satisfies the equality $\operatorname{sen}\left(x + \dfrac{\pi}{4}\right) + \operatorname{sen}\left(x - \dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2}$.
Ver resposta
$S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{\pi}{6} + 2k\pi \text{ ou } x = \dfrac{5\pi}{6} + 2k\pi\right\}$
(MAUÁ-77) Given the system $\begin{cases} \operatorname{sen}(x + y) + \operatorname{sen}(x - y) = 2 \\ \operatorname{sen} x + \cos y = 2 \end{cases}$: a) show that the pair $(x_0, y_0)$, with $x_0 = 2\pi$ and $y_0 = \dfrac{\pi}{2}$, is not a solution of the system; b) solve the system, determining all solutions $(x, y)$.
Ver resposta
a) $2\operatorname{sen}(2\pi)\cos\frac{\pi}{2} = 0 \ne 2$ · b) $x = \dfrac{\pi}{2} + 2k\pi$ e $y = 2k\pi$
(FEI-77) Solve the system $\begin{cases} \operatorname{sen} a + \cos b = 1 \\ \operatorname{sen}\dfrac{a + b}{2} \cdot \cos\dfrac{a - b}{2} = \dfrac{1}{2} \end{cases}$, with $a$ and $b$ from the 1st quadrant.
Ver resposta
$a = \operatorname{arc\,sen}\dfrac{2 - \sqrt{2}}{2}$ e $b = \dfrac{\pi}{4}$
Solve the equation $\operatorname{sen}^4 x + \cos^4 x = \dfrac{3}{4}$.
See the resolution
It follows from the theory that $\operatorname{sen}^2 2x = 2\left(1 - \dfrac{3}{4}\right) = \dfrac{1}{2}$, therefore $\operatorname{sen} 2x = \pm\dfrac{\sqrt{2}}{2}$ and then $2x = \dfrac{\pi}{4} + \dfrac{k\pi}{2} \Rightarrow x = \dfrac{\pi}{8} + \dfrac{k\pi}{4}$.
$$S = \left\{x \in \mathbb{R} \;\middle|\; x = \frac{\pi}{8} + \frac{k\pi}{4}\right\}$$Solve the equation $\operatorname{sen}^6 x + \cos^6 x = \dfrac{7}{16}$.
See the resolution
It follows from the theory that $\operatorname{sen}^2 2x = \dfrac{4}{3}(1 - a) = \dfrac{4}{3}\left(1 - \dfrac{7}{16}\right) = \dfrac{3}{4}$, therefore $\operatorname{sen} 2x = \pm\dfrac{\sqrt{3}}{2}$ and then $2x = \pm\dfrac{\pi}{3} + k\pi \Rightarrow x = \pm\dfrac{\pi}{6} + \dfrac{k\pi}{2}$.
$$S = \left\{x \in \mathbb{R} \;\middle|\; x = \pm\frac{\pi}{6} + \frac{k\pi}{2}\right\}$$Solve the equations for $x \in [0, 2\pi]$: a) $\operatorname{sen}^4 x + \cos^4 x = \dfrac{5}{8}$; b) $\operatorname{sen}^6 x + \cos^6 x = \dfrac{5}{8}$; c) $\operatorname{sen}^4 x + \cos^4 x = \dfrac{1}{2}$; d) $\operatorname{sen}^6\dfrac{x}{2} + \cos^6\dfrac{x}{2} = \dfrac{7}{16}$; e) $\operatorname{sen}^3 x + \cos^3 x = 1$.
Ver resposta
a) $S = \left\{\dfrac{\pi}{6}, \dfrac{\pi}{3}, \dfrac{2\pi}{3}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{4\pi}{3}, \dfrac{5\pi}{3}, \dfrac{11\pi}{6}\right\}$ · b) $S = \left\{\dfrac{\pi}{8}, \dfrac{3\pi}{8}, \dfrac{5\pi}{8}, \dfrac{7\pi}{8}, \dfrac{9\pi}{8}, \dfrac{11\pi}{8}, \dfrac{13\pi}{8}, \dfrac{15\pi}{8}\right\}$ · c) $S = \left\{\dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{7\pi}{4}\right\}$ · d) $S = \left\{\dfrac{\pi}{3}, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}, \dfrac{5\pi}{3}\right\}$ · e) $S = \left\{0, \dfrac{\pi}{2}, 2\pi\right\}$
Inverse circular functions
Determine $\alpha$ such that $\alpha = \operatorname{arc\,sen}\dfrac{1}{2}$.
See the resolution
$\alpha = \operatorname{arc\,sen}\dfrac{1}{2} \iff \operatorname{sen}\alpha = \dfrac{1}{2}$ and $-\dfrac{\pi}{2} \le \alpha \le \dfrac{\pi}{2}$. That is, $\operatorname{arc\,sen}\frac{1}{2}$ is not any $\alpha$ such as $\operatorname{sen}\alpha = \frac{1}{2}$, but that (single) $\alpha$ that is in the range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$: $\alpha = \dfrac{\pi}{6}$.
Determine the numbers: $\operatorname{arc\,sen} 0$, $\operatorname{arc\,sen}\dfrac{\sqrt{3}}{2}$, $\operatorname{arc\,sen}\left(-\dfrac{1}{2}\right)$, $\operatorname{arc\,sen} 1$ and $\operatorname{arc\,sen}(-1)$.
Ver resposta
$0$, $\dfrac{\pi}{3}$, $-\dfrac{\pi}{6}$, $\dfrac{\pi}{2}$ e $-\dfrac{\pi}{2}$
Calculate $\cos\left(\operatorname{arc\,sen}\dfrac{1}{3}\right)$.
See the resolution
Doing $\operatorname{arc\,sen}\frac{1}{3} = \alpha$, we have $\operatorname{sen}\alpha = \frac{1}{3}$ and $-\frac{\pi}{2} \le \alpha \le \frac{\pi}{2}$; so the cosine is positive and
$$\cos\alpha = +\sqrt{1 - \operatorname{sen}^2\alpha} = +\sqrt{1 - \frac{1}{9}} = +\sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}.$$Calculate $\operatorname{tg}\left(\operatorname{arc\,sen}\dfrac{3}{4}\right)$.
Ver resposta
$\operatorname{tg}\left(\operatorname{arc\,sen}\dfrac{3}{4}\right) = \dfrac{3\sqrt{7}}{7}$
Calculate $\cos\left(\operatorname{arc\,sen}\dfrac{3}{5} + \operatorname{arc\,sen}\dfrac{5}{13}\right)$.
See the resolution
Doing $\operatorname{arc\,sen}\frac{3}{5} = \alpha$: $\operatorname{sen}\alpha = \frac{3}{5}$, $-\frac{\pi}{2} \le \alpha \le \frac{\pi}{2}$, then $\cos\alpha = +\sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5}$. Doing $\operatorname{arc\,sen}\frac{5}{13} = \beta$: $\operatorname{sen}\beta = \frac{5}{13}$, then $\cos\beta = +\sqrt{1 - \left(\frac{5}{13}\right)^2} = \frac{12}{13}$. Finally,
$$\cos(\alpha + \beta) = \cos\alpha\cos\beta - \operatorname{sen}\alpha\operatorname{sen}\beta = \frac{4}{5} \cdot \frac{12}{13} - \frac{3}{5} \cdot \frac{5}{13} = \frac{48 - 15}{65} = \frac{33}{65}.$$Calculate: a) $\operatorname{tg}\left(\operatorname{arc\,sen}\left(-\dfrac{2}{3}\right) + \operatorname{arc\,sen}\dfrac{1}{4}\right)$; b) $\operatorname{sen}\left(2\operatorname{arc\,sen}\left(-\dfrac{3}{5}\right)\right)$; c) $\cos\left(3\operatorname{arc\,sen}\dfrac{12}{13}\right)$.
Ver resposta
a) $\dfrac{\sqrt{5}(-6 + \sqrt{3})}{15 + 2\sqrt{3}}$ · b) $-\dfrac{24}{25}$ · c) $-\dfrac{2035}{2197}$
Determine $\alpha$ such that $\alpha = \operatorname{arc\,cos}\dfrac{\sqrt{3}}{2}$.
See the resolution
$\alpha = \operatorname{arc\,cos}\dfrac{\sqrt{3}}{2} \Rightarrow \cos\alpha = \dfrac{\sqrt{3}}{2}$ and $0 \le \alpha \le \pi$; then $\alpha = \dfrac{\pi}{6}$.
Determine the numbers: $\operatorname{arc\,cos} 1$, $\operatorname{arc\,cos}\dfrac{1}{2}$, $\operatorname{arc\,cos}\dfrac{\sqrt{2}}{2}$, $\operatorname{arc\,cos} 0$ and $\operatorname{arc\,cos}(-1)$.
Ver resposta
$0$, $\dfrac{\pi}{3}$, $\dfrac{\pi}{4}$, $\dfrac{\pi}{2}$ e $\pi$
Calculate $\operatorname{tg}\left(\operatorname{arc\,cos}\dfrac{2}{5}\right)$.
See the resolution
Making $\operatorname{arc\,cos}\frac{2}{5} = \alpha$: $\cos\alpha = \frac{2}{5}$ and $0 \le \alpha \le \pi$, then $\operatorname{sen}\alpha = +\sqrt{1 - \cos^2\alpha} = +\sqrt{1 - \frac{4}{25}} = \frac{\sqrt{21}}{5}$ and $\operatorname{tg}\alpha = \dfrac{\operatorname{sen}\alpha}{\cos\alpha} = \dfrac{\sqrt{21}}{2}$.
Calculate $\operatorname{sen}\left(\operatorname{arc\,cos}\left(-\dfrac{3}{5}\right)\right)$.
Ver resposta
$\operatorname{sen}\left(\operatorname{arc\,cos}\left(-\dfrac{3}{5}\right)\right) = \dfrac{4}{5}$
Calculate $\operatorname{cotg}\left(\operatorname{arc\,cos}\dfrac{2}{7}\right)$.
Ver resposta
$\dfrac{2\sqrt{5}}{15}$
(FEI-76) Being $A$ from the first quadrant and $\operatorname{arc\,sen} x = A$, find $\operatorname{arc\,cos} x$.
Ver resposta
$\operatorname{arc\,cos} x = \dfrac{\pi}{2} - A$
Calculate $\operatorname{sen}\left(\operatorname{arc\,cos}\dfrac{5}{13} + \operatorname{arc\,sen}\dfrac{7}{25}\right)$.
See the resolution
Doing $\operatorname{arc\,cos}\frac{5}{13} = \alpha$: $\cos\alpha = \frac{5}{13}$ and $0 \le \alpha \le \pi$, then $\operatorname{sen}\alpha = +\sqrt{1 - \left(\frac{5}{13}\right)^2} = \frac{12}{13}$. Doing $\operatorname{arc\,sen}\frac{7}{25} = \beta$: $\operatorname{sen}\beta = \frac{7}{25}$ and $-\frac{\pi}{2} \le \beta \le \frac{\pi}{2}$, then $\cos\beta = \frac{24}{25}$. Finally,
$$\operatorname{sen}(\alpha + \beta) = \operatorname{sen}\alpha\cos\beta + \operatorname{sen}\beta\cos\alpha = \frac{12}{13} \cdot \frac{24}{25} + \frac{7}{25} \cdot \frac{5}{13} = \frac{288 + 35}{325} = \frac{323}{325}.$$Calculate: a) $\operatorname{sen}\left(\operatorname{arc\,cos}\dfrac{3}{5} - \operatorname{arc\,cos}\dfrac{5}{13}\right)$; b) $\cos\left(\operatorname{arc\,sen}\dfrac{7}{25} - \operatorname{arc\,cos}\dfrac{12}{13}\right)$; c) $\operatorname{tg}\left(2\operatorname{arc\,cos}\left(-\dfrac{3}{5}\right)\right)$; d) $\cos\left(\dfrac{1}{2}\operatorname{arc\,cos}\dfrac{7}{25}\right)$.
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a) $-\dfrac{16}{65}$ · b) $\dfrac{323}{325}$ · c) $\dfrac{24}{7}$ · d) $\dfrac{4}{5}$
(MAPOFEI-72) Let the function be $f(x) = \cos(2\operatorname{arc\,cos} x)$, $-1 \le x \le 1$. a) Determine the values of $x$ such that $f(x) = 0$. b) Sketch the graph of $g(x) = \dfrac{1}{f(x)}$.
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With $\alpha = \operatorname{arc\,cos} x$, $f(x) = \cos 2\alpha = 2\cos^2\alpha - 1 = 2x^2 - 1$.
a) $2x^2 - 1 = 0 \Rightarrow x = \pm\dfrac{\sqrt{2}}{2}$.
b) $g(x) = \dfrac{1}{2x^2 - 1}$ in $[-1, 1]$, except $x = \pm\frac{\sqrt{2}}{2}$: is worth 1 in $x = \pm 1$, is worth $-1$ in $x = 0$ and has vertical asymptotes in $x = \pm\frac{\sqrt{2}}{2}$.
Determine $\alpha$ such that $\alpha = \operatorname{arc\,tg} 1$.
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$\alpha = \operatorname{arc\,tg} 1 \iff \operatorname{tg}\alpha = 1$ and $-\dfrac{\pi}{2} < \alpha < \dfrac{\pi}{2}$, that is, $\alpha = \dfrac{\pi}{4}$.
Determine the numbers: $\operatorname{arc\,tg} 0$, $\operatorname{arc\,tg}\sqrt{3}$, $\operatorname{arc\,tg}(-1)$ and $\operatorname{arc\,tg}\left(-\dfrac{\sqrt{3}}{3}\right)$.
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$0$, $\dfrac{\pi}{3}$, $-\dfrac{\pi}{4}$ e $-\dfrac{\pi}{6}$
Calculate $\operatorname{sen}(\operatorname{arc\,tg}\sqrt{2})$.
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Making $\operatorname{arc\,tg}\sqrt{2} = \alpha$: $\operatorname{tg}\alpha = \sqrt{2}$ and $-\frac{\pi}{2} < \alpha < \frac{\pi}{2}$ (actually $\alpha > 0$). Then
$$\operatorname{sen}^2\alpha = \frac{\operatorname{tg}^2\alpha}{1 + \operatorname{tg}^2\alpha} = \frac{2}{1 + 2} = \frac{2}{3} \;\Longrightarrow\; \operatorname{sen}\alpha = +\sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}.$$Calculate $\cos\left(\operatorname{arc\,tg}\left(-\dfrac{4}{3}\right)\right)$.
Ver resposta
$\dfrac{3}{5}$
Calculate $\operatorname{tg}\left(\operatorname{arc\,sen}\dfrac{3}{5} - \operatorname{arc\,tg}\dfrac{5}{12}\right)$.
See the resolution
Doing $\operatorname{arc\,sen}\frac{3}{5} = \alpha$: $\operatorname{sen}\alpha = \frac{3}{5}$, $-\frac{\pi}{2} \le \alpha \le \frac{\pi}{2}$, then $\cos\alpha = \frac{4}{5}$ and $\operatorname{tg}\alpha = \frac{3}{4}$. Doing $\operatorname{arc\,tg}\frac{5}{12} = \beta$: $\operatorname{tg}\beta = \frac{5}{12}$. Finally,
$$\operatorname{tg}(\alpha - \beta) = \frac{\operatorname{tg}\alpha - \operatorname{tg}\beta}{1 + \operatorname{tg}\alpha\operatorname{tg}\beta} = \frac{\frac{3}{4} - \frac{5}{12}}{1 + \frac{3}{4} \cdot \frac{5}{12}} = \frac{\frac{16}{48}}{\frac{63}{48}} = \frac{16}{63}.$$Calculate: a) $\operatorname{sen}(\operatorname{arc\,tg} 2 + \operatorname{arc\,tg} 3)$; b) $\cos\left(\operatorname{arc\,tg} 2 - \operatorname{arc\,tg}\dfrac{1}{2}\right)$; c) $\operatorname{tg}\left(2\operatorname{arc\,tg}\dfrac{1}{5}\right)$; d) $\cos\left(3\operatorname{arc\,tg}\dfrac{24}{7}\right)$.
Ver resposta
a) $\dfrac{\sqrt{2}}{2}$ · b) $\dfrac{4}{5}$ · c) $\dfrac{5}{12}$ · d) $-\dfrac{11\,753}{15\,625}$. (In a and b, the printed template shows $\pm$; but the arcs have well-defined values: in a, $\operatorname{arc\,tg} 2 + \operatorname{arc\,tg} 3 = \frac{3\pi}{4}$, whose sine is positive; in b, the difference is in $\left]0, \frac{\pi}{2}\right[$ and the cosine is positive.)
Demonstrate equality $\operatorname{arc\,sen}\dfrac{\sqrt{5}}{5} + \operatorname{arc\,cos}\dfrac{3}{\sqrt{10}} = \operatorname{arc\,tg} 1$.
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I) Let’s do $\alpha = \operatorname{arc\,sen}\frac{\sqrt{5}}{5}$, $\beta = \operatorname{arc\,cos}\frac{3}{\sqrt{10}}$ and $\gamma = \operatorname{arc\,tg} 1$. Then:
- $\operatorname{sen}\alpha = \frac{\sqrt{5}}{5}$ and $-\frac{\pi}{2} \le \alpha \le \frac{\pi}{2} \Rightarrow 0 < \alpha < \frac{\pi}{4}$ (because $\frac{\sqrt{5}}{5} < \frac{\sqrt{2}}{2}$);
- $\cos\beta = \frac{3}{\sqrt{10}}$ and $0 \le \beta \le \pi \Rightarrow 0 < \beta < \frac{\pi}{4}$ (because $\frac{3}{\sqrt{10}} > \frac{\sqrt{2}}{2}$);
then $0 < \alpha + \beta < \frac{\pi}{2}$. And $\operatorname{tg}\gamma = 1$, $-\frac{\pi}{2} < \gamma < \frac{\pi}{2} \Rightarrow \gamma = \frac{\pi}{4}$.
II) Let us calculate $\operatorname{tg}(\alpha + \beta)$. Like $\cos\alpha = \frac{\sqrt{20}}{5}$ and $\operatorname{sen}\beta = \frac{1}{\sqrt{10}}$, we have $\operatorname{tg}\alpha = \frac{1}{2}$ and $\operatorname{tg}\beta = \frac{1}{3}$:
$$\operatorname{tg}(\alpha + \beta) = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \cdot \frac{1}{3}} = \frac{\frac{5}{6}}{\frac{5}{6}} = 1 = \operatorname{tg}\gamma.$$III) Conclusion: $\alpha + \beta$ and $\gamma$ are both in $\left]0, \frac{\pi}{2}\right[$, where the tangent is one-to-one, and have the same tangent; logo $\alpha + \beta = \gamma$.
Prove the equalities: a) (Itajubá-77) $\operatorname{arc\,tg}\dfrac{1}{2} + \operatorname{arc\,tg}\dfrac{1}{3} = \dfrac{\pi}{4}$; b) $\operatorname{arc\,sen}\dfrac{1}{\sqrt{5}} + \operatorname{arc\,cos}\dfrac{3}{\sqrt{10}} = \dfrac{\pi}{4}$; c) $\operatorname{arc\,cos}\dfrac{3}{5} + \operatorname{arc\,cos}\dfrac{12}{13} = \operatorname{arc\,cos}\dfrac{16}{65}$; d) $\operatorname{arc\,sen}\dfrac{24}{25} - \operatorname{arc\,sen}\dfrac{3}{5} = \operatorname{arc\,tg}\dfrac{3}{4}$.
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In all items we followed the exercise guide 267: finding the arcs, calculating a sum function and using injectivity.
a) $\alpha = \operatorname{arc\,tg}\frac{1}{2}$ and $\beta = \operatorname{arc\,tg}\frac{1}{3}$ are in $\left]0, \frac{\pi}{4}\right[$, therefore $0 < \alpha + \beta < \frac{\pi}{2}$; and $\operatorname{tg}(\alpha + \beta) = \dfrac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{6}} = 1$. Then $\alpha + \beta = \frac{\pi}{4}$.
b) $\operatorname{arc\,sen}\frac{1}{\sqrt{5}}$ has tangent $\frac{1}{2}$ and $\operatorname{arc\,cos}\frac{3}{\sqrt{10}}$ has tangent $\frac{1}{3}$, both in the 1st quadrant: it is item (a) (and exercise 267).
c) $\alpha = \operatorname{arc\,cos}\frac{3}{5}$ ($\operatorname{sen}\alpha = \frac{4}{5}$) and $\beta = \operatorname{arc\,cos}\frac{12}{13}$ ($\operatorname{sen}\beta = \frac{5}{13}$) are in $\left]0, \frac{\pi}{2}\right[$, therefore $\alpha + \beta \in\, ]0, \pi[$, where the cosine is injective; and $\cos(\alpha + \beta) = \frac{3}{5} \cdot \frac{12}{13} - \frac{4}{5} \cdot \frac{5}{13} = \frac{16}{65}$. Then $\alpha + \beta = \operatorname{arc\,cos}\frac{16}{65}$.
d) $\alpha = \operatorname{arc\,sen}\frac{24}{25}$ ($\operatorname{tg}\alpha = \frac{24}{7}$) and $\beta = \operatorname{arc\,sen}\frac{3}{5}$ ($\operatorname{tg}\beta = \frac{3}{4}$), with $0 < \beta < \alpha < \frac{\pi}{2}$, therefore $0 < \alpha - \beta < \frac{\pi}{2}$; and $\operatorname{tg}(\alpha - \beta) = \dfrac{\frac{24}{7} - \frac{3}{4}}{1 + \frac{24}{7} \cdot \frac{3}{4}} = \dfrac{\frac{75}{28}}{\frac{100}{28}} = \dfrac{3}{4}$. Then $\alpha - \beta = \operatorname{arc\,tg}\frac{3}{4}$.
Prove that $2\operatorname{arc\,tg}\dfrac{1}{3} + \operatorname{arc\,tg}\dfrac{1}{7} = \dfrac{\pi}{4}$.
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Making $\operatorname{arc\,tg}\frac{1}{3} = \alpha$: $\operatorname{tg}\alpha = \frac{1}{3}$ and $-\frac{\pi}{2} < \alpha < \frac{\pi}{2} \Rightarrow 0 < \alpha < \frac{\pi}{4} \Rightarrow 0 < 2\alpha < \frac{\pi}{2}$ (1). Making $\operatorname{arc\,tg}\frac{1}{7} = \beta$: $\operatorname{tg}\beta = \frac{1}{7}$ and $0 < \beta < \frac{\pi}{2}$ (2).
Considering (1) and (2), $0 < 2\alpha + \beta < \pi$, and to prove that $2\alpha + \beta = \frac{\pi}{4}$ it is enough to prove that $\operatorname{tg}(2\alpha + \beta) = 1$:
$$\operatorname{tg} 2\alpha = \frac{2\operatorname{tg}\alpha}{1 - \operatorname{tg}^2\alpha} = \frac{\frac{2}{3}}{1 - \frac{1}{9}} = \frac{3}{4} \qquad \operatorname{tg}(2\alpha + \beta) = \frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{4} \cdot \frac{1}{7}} = \frac{\frac{25}{28}}{\frac{25}{28}} = 1.$$Prove the equalities: a) $2\operatorname{arc\,tg}\dfrac{2}{3} + \operatorname{arc\,cos}\dfrac{12}{13} = \dfrac{\pi}{2}$; b) $3\operatorname{arc\,sen}\dfrac{1}{4} + \operatorname{arc\,cos}\dfrac{11}{16} = \dfrac{\pi}{2}$.
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a) Let $\alpha = \operatorname{arc\,tg}\frac{2}{3}$ be: $0 < \alpha < \frac{\pi}{4}$, then $0 < 2\alpha < \frac{\pi}{2}$, and $\operatorname{tg} 2\alpha = \dfrac{\frac{4}{3}}{1 - \frac{4}{9}} = \dfrac{12}{5}$. Let $\beta = \operatorname{arc\,cos}\frac{12}{13}$: $0 < \beta < \frac{\pi}{2}$ and $\operatorname{cotg}\beta = \frac{12}{5}$. Then $\operatorname{tg} 2\alpha = \operatorname{cotg}\beta = \operatorname{tg}\left(\frac{\pi}{2} - \beta\right)$, with $2\alpha$ and $\frac{\pi}{2} - \beta$ in $\left]0, \frac{\pi}{2}\right[$; logo $2\alpha = \frac{\pi}{2} - \beta$.
b) Let $\alpha = \operatorname{arc\,sen}\frac{1}{4}$: as $\frac{1}{4} < \frac{1}{2}$, $0 < \alpha < \frac{\pi}{6}$ and $0 < 3\alpha < \frac{\pi}{2}$. By the triple arc, $\operatorname{sen} 3\alpha = 3 \cdot \frac{1}{4} - 4 \cdot \frac{1}{64} = \frac{11}{16}$. Let $\beta = \operatorname{arc\,cos}\frac{11}{16} \in \left]0, \frac{\pi}{2}\right[$: $\cos\beta = \frac{11}{16} = \operatorname{sen} 3\alpha = \cos\left(\frac{\pi}{2} - 3\alpha\right)$. As $\beta$ and $\frac{\pi}{2} - 3\alpha$ are in $\left]0, \frac{\pi}{2}\right[$, $\beta = \frac{\pi}{2} - 3\alpha$.