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Tests — Inverse circular functions

Arc sine, arc cosine and arc tangent (chapter VII). Teste 158MACK-74 Let $f$, $g$ and $h$ be functions of $A$ in $A$, where $A = [-1, 1]$, defined as follows: $f(x) = \operatorname{sen} x$, $g(x) = \operatorname{sen}\pi x$, $h(x) = \frac{\pi}{2}x$. We can state that: a) they are all invertibleb) they are all superjectivec) only one is injectabled) only one is surjectivee) only one is injective and surjective $f$ (sine in $[-1, 1]$) is injective, with image $[-\operatorname{sen} 1, \operatorname{sen} 1]$; $g(x) = \operatorname{sen}\pi x$ has image $[-1, 1]$ (surjective), but is not injective ($g(0) = g(1) = 0$); $h$ is injective and its image $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ doesn’t even fit into $A$. Only $g$ is surjective. The printed template indicates (e), but none of the three is both an injector and a surjector of $A$ in $A$; the true statement is (d). Teste 159MACK-73 The domain of the function defined by $y = \operatorname{arc\,sen}\sqrt{2x - 3}$ is: a) $\left\{x \in \mathbb{R} \;\middle|\; x \ge \frac{3}{2}\right\}$b) $\left\{x \in \mathbb{R} \;\middle|\; \frac{3}{2} \le x \le 2\right\}$c) $\{x \in \mathbb{R} \mid 0 \le x \le 2\}$d) $\left\{x \in \mathbb{R} \;\middle|\; -2 \le x \le 0 \text{ ou } \frac{5}{2} \le x \le 4\right\}$e) none of the previous answers The root requires $2x - 3 \ge 0$ and the arcsine requires $\sqrt{2x - 3} \le 1$: $0 \le 2x - 3 \le 1$, that is, $\frac{3}{2} \le x \le 2$.

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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) $\operatorname{sen} x = -1 = \operatorname{sen}\dfrac{3\pi}{2}$. $\;S = \left\{x \in \mathbb{R} \;\middle|\; x = \dfrac{3\pi}{2} + 2k\pi\right\}$.