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Tests — Equations

Trigonometric equations and solutions on an interval (chapter VII). Teste 122FUVEST-77 In the range $\frac{\pi}{2} \le x \le \pi$, the equation $\sqrt{1 - \operatorname{sen}^2 x} + \cos x = -\sqrt{2}$: a) does not admit solutionb) admits as a solution $x = \frac{3\pi}{4}$c) admits as a solution $x = \frac{2\pi}{3}$d) admits as a solution $x = \frac{5\pi}{6}$e) admits as a solution $x = \pi$ $\sqrt{1 - \operatorname{sen}^2 x} = |\cos x|$, and in the 2nd quadrant $|\cos x| = -\cos x$. The first member is worth $-\cos x + \cos x = 0 \ne -\sqrt{2}$. Teste 123PUC-76 The values ​​of $x$ that satisfy the equation $\cos\left(3x - \frac{\pi}{5}\right) = 0$ are ($k = 0, \pm 1, \pm 2, \ldots$): a) $x = \frac{7\pi}{30} + k\frac{\pi}{3}$b) $x = \frac{7\pi}{15} + k\frac{\pi}{3}$c) $x = \frac{7\pi}{2} + k\frac{\pi}{4}$d) $x = \frac{7\pi}{5} + k\frac{\pi}{2}$e) $x = \frac{7\pi}{4} + k\frac{\pi}{6}$ $3x - \frac{\pi}{5} = \frac{\pi}{2} + k\pi \Rightarrow 3x = \frac{7\pi}{10} + k\pi \Rightarrow x = \frac{7\pi}{30} + k\frac{\pi}{3}$.

October 25, 2026 Read
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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 193 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 resolutiona) $\operatorname{sen} x = \operatorname{sen}\dfrac{\pi}{5} \Rightarrow x = \dfrac{\pi}{5} + 2k\pi$ 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) $\operatorname{sen} x = -1 = \operatorname{sen}\dfrac{3\pi}{2}$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{3\pi}{2} + 2k\pi\right\}$.

October 24, 2026 Read
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Inverse circular functions

The equation $\operatorname{sen} x = \frac{1}{3}$ has infinitely many solutions, but none of them is a notable arc. How to write the solution? We need a name for “the arc whose sine is $\frac{1}{3}$”. The problem is that there are infinitely many arcs like this; so that the name designates a number, we choose an interval where the sine is not repeated. This choice defines the inverse circular functions. The story The notation $\operatorname{arc\,sen} x$ — “the arc whose sine is $x$” — was used by Daniel Bernoulli in 1729 and by the Austrian mathematician Joseph Scherffer in 1772, and gained traction with Lagrange and other authors from the end of the 18th century. The notation $\operatorname{sen}^{-1} x$, common in calculators, was proposed by the English astronomer John Herschel in 1813, by analogy with the inverse of a function; it has the defect of being confused with $\frac{1}{\operatorname{sen} x}$, which is why in this book we use $\operatorname{arc\,sen}$.

October 24, 2026 Read
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Solutions on an interval and classical equations

So far we have solved the equations in $\mathbb{R}$, obtaining infinite families of solutions. Often, however, only solutions in a range are of interest — for example, the angles of a triangle, which are between $0$ and $\pi$. And certain equations appear so much that they have their own methods: they are classical equations. The story The equation $a\operatorname{sen} x + b\cos x = c$ is the equation of the sum of two oscillations of the same period: it appears in astronomy (the combination of two circular movements), in wave physics and in electricity (out-of-phase alternating currents). The trick of rewriting $a\operatorname{sen} x + b\cos x$ as a single displaced sinusoid — the auxiliary angle $\theta$ — was already used by 18th century astronomers. It gained industrial importance at the end of the 19th century, when the engineer Charles Steinmetz showed, in 1893, that alternating currents could be added as complex numbers: each sinusoid becomes an arrow with length (the amplitude) and slope (the phase), and adding oscillations becomes adding arrows. This is exactly what we do when writing $a\operatorname{sen} x + b\cos x = \sqrt{a^2 + b^2}\,\operatorname{sen}(x + \theta)$.

October 23, 2026 Read
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Fundamental equations

Let $f$ and $g$ be two trigonometric functions of the variable $x$. Solving the equation $f(x) = g(x)$ is determining the set $S$ of numbers $r$ for which $f(r) = g(r)$ is a true sentence. This set is the solution set (or truth set) of the equation. The story Trigonometric equations appeared long before they had that name: Greek, Indian and Arab astronomers needed to find the arc that has a given sine to convert observations into positions in the sky. The most famous of these is the Kepler equation, published in Astronomia nova (1609):

October 23, 2026 Read