Trigonometria
31 Equações e Inequações Trigonométricas
Resumo do capítulo
Encerrando o livro, este capítulo reúne as técnicas dos capítulos anteriores -- círculo unitário, identidades e propriedades das funções inversas -- para resolver equações e inequações trigonométricas, expressando o conjunto-solução em termos de todos os ângulos côngruos.
Nos Exercícios 1 a 18, encontre todas as soluções exatas da equação e, em seguida, liste aquelas que estão no intervalo $[0, 2\pi)$.
- $x = \dfrac{\pi k}{5}; \; x = 0, \dfrac{\pi}{5}, \dfrac{2\pi}{5}, \dfrac{3\pi}{5}, \dfrac{4\pi}{5}, \pi, \dfrac{6\pi}{5}, \dfrac{7\pi}{5}, \dfrac{8\pi}{5}, \dfrac{9\pi}{5}$
- $\cos \left( 3x \right) = \dfrac{1}{2}$resposta$x = \dfrac{\pi}{9} + \dfrac{2\pi k}{3}$ ou $x = \dfrac{5\pi}{9} + \dfrac{2\pi k}{3}; \; x = \dfrac{\pi}{9}, \dfrac{5\pi}{9}, \dfrac{7\pi}{9}, \dfrac{11\pi}{9}, \dfrac{13\pi}{9}, \dfrac{17\pi}{9}$
- $\sin \left( -2x \right) = \dfrac{\sqrt{3}}{2}$resposta$x = \dfrac{2\pi}{3} + \pi k$ ou $x = \dfrac{5\pi}{6} + \pi k; \; x = \dfrac{2\pi}{3}, \dfrac{5\pi}{6}, \dfrac{5\pi}{3}, \dfrac{11\pi}{6}$
- $\tan \left( 6x \right) = 1$resposta$x = \dfrac{\pi}{24} + \dfrac{\pi k}{6}; \; x = \dfrac{\pi}{24}, \dfrac{5\pi}{24}, \dfrac{3\pi}{8}, \dfrac{13\pi}{24}, \dfrac{17\pi}{24}, \dfrac{7\pi}{8}, \dfrac{25\pi}{24}, \dfrac{29\pi}{24}, \dfrac{11\pi}{8}, \dfrac{37\pi}{24}, \dfrac{41\pi}{24}, \dfrac{15\pi}{8}$
- $\csc \left( 4x \right) = -1$resposta$x = \dfrac{3\pi}{8} + \dfrac{\pi k}{2}; \; x = \dfrac{3\pi}{8}, \dfrac{7\pi}{8}, \dfrac{11\pi}{8}, \dfrac{15\pi}{8}$
- $\sec \left( 3x \right) = \sqrt{2}$resposta$x = \dfrac{\pi}{12} + \dfrac{2\pi k}{3}$ ou $x = \dfrac{7\pi}{12} + \dfrac{2\pi k}{3}; \; x = \dfrac{\pi}{12}, \dfrac{7\pi}{12}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{17\pi}{12}, \dfrac{23\pi}{12}$
- $\cot \left( 2x \right) = -\dfrac{\sqrt{3}}{3}$resposta$x = \dfrac{\pi}{3} + \dfrac{\pi k}{2}; \; x = \dfrac{\pi}{3}, \dfrac{5\pi}{6}, \dfrac{4\pi}{3}, \dfrac{11\pi}{6}$
- $\cos \left( 9x \right) = 9$respostaSem solução
- $\sin \left( \dfrac{x}{3} \right) = \dfrac{\sqrt{2}}{2}$resposta$x = \dfrac{3\pi}{4} + 6\pi k$ ou $x = \dfrac{9\pi}{4} + 6\pi k; \; x = \dfrac{3\pi}{4}$
- $\cos \left( x + \dfrac{5\pi}{6} \right) = 0$resposta$x = -\dfrac{\pi}{3} + \pi k; \; x = \dfrac{2\pi}{3}, \dfrac{5\pi}{3}$
- $\sin \left( 2x - \dfrac{\pi}{3} \right) = -\dfrac{1}{2}$resposta$x = \dfrac{3\pi}{4} + \pi k$ ou $x = \dfrac{13\pi}{12} + \pi k; \; x = \dfrac{\pi}{12}, \dfrac{3\pi}{4}, \dfrac{13\pi}{12}, \dfrac{7\pi}{4}$
- $2\cos \left( x + \dfrac{7\pi}{4} \right) = \sqrt{3}$resposta$x = -\dfrac{19\pi}{12} + 2\pi k$ ou $x = \dfrac{\pi}{12} + 2\pi k; \; x = \dfrac{\pi}{12}, \dfrac{5\pi}{12}$
- $\csc(x) = 0$respostaSem solução
- $\tan \left( 2x - \pi \right) = 1$resposta$x = \dfrac{5\pi}{8} + \dfrac{\pi k}{2}; \; x = \dfrac{\pi}{8}, \dfrac{5\pi}{8}, \dfrac{9\pi}{8}, \dfrac{13\pi}{8}$
- $\tan^{2} \left( x \right) = 3$resposta$x = \dfrac{\pi}{3} + \pi k$ ou $x = \dfrac{2\pi}{3} + \pi k; \; x = \dfrac{\pi}{3}, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}, \dfrac{5\pi}{3}$
- $\sec^{2} \left( x \right) = \dfrac{4}{3}$resposta$x = \dfrac{\pi}{6} + \pi k$ ou $x = \dfrac{5\pi}{6} + \pi k; \; x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$
- $\cos^{2} \left( x \right) = \dfrac{1}{2}$resposta$x = \dfrac{\pi}{4} + \dfrac{\pi k}{2}; \; x = \dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{7\pi}{4}$
- $x = \dfrac{\pi}{3} + \pi k$ ou $x = \dfrac{2\pi}{3} + \pi k; \; x = \dfrac{\pi}{3}, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}, \dfrac{5\pi}{3}$
Nos Exercícios 19 a 42, resolva a equação, fornecendo as soluções exatas que estão no intervalo $[0, 2\pi)$.
- $x = \dfrac{\pi}{4}, \dfrac{5\pi}{4}$
- $\sin \left( 2x \right) = \sin \left( x \right)$resposta$x = 0, \dfrac{\pi}{3}, \pi, \dfrac{5\pi}{3}$
- $\sin \left( 2x \right) = \cos \left( x \right)$resposta$x = \dfrac{\pi}{6}, \dfrac{\pi}{2}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}$
- $\cos \left( 2x \right) = \sin \left( x \right)$resposta$x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}$
- $\cos \left( 2x \right) = \cos \left( x \right)$resposta$x = 0, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}$
- $\cos(2x) = 2 - 5\cos(x)$resposta$x=\dfrac{\pi}{3}, \dfrac{5\pi}{3}$
- $3\cos(2x) + \cos(x) + 2 = 0$resposta$x = \dfrac{2\pi}{3}, \dfrac{4\pi}{3}, \arccos\left(\dfrac{1}{3}\right), 2\pi -\arccos\left(\dfrac{1}{3}\right) $
- $\cos(2x) = 5\sin(x) - 2$resposta$x=\dfrac{\pi}{6}, \dfrac{5\pi}{6}$
- $3\cos(2x) = \sin(x) + 2$resposta$x = \dfrac{7\pi}{6}, \dfrac{11\pi}{6}, \arcsin\left(\dfrac{1}{3}\right), \pi - \arcsin\left(\dfrac{1}{3}\right) $
- $2\sec^{2}(x) = 3 - \tan(x)$resposta$x=\dfrac{3\pi}{4}, \dfrac{7\pi}{4}, \arctan\left(\dfrac{1}{2}\right), \pi +\arctan\left(\dfrac{1}{2}\right) $
- $\tan^{2}(x) = 1-\sec(x)$resposta$x=0, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}$
- $\cot^{2}(x) = 3\csc(x) - 3$resposta$x=\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{\pi}{2}$
- $\sec(x) = 2\csc(x)$resposta$x=\arctan(2), \pi + \arctan(2)$
- $\cos(x)\csc(x)\cot(x) = 6-\cot^{2}(x)$resposta$x = \dfrac{\pi}{6}, \dfrac{7\pi}{6}, \dfrac{5\pi}{6}, \dfrac{11\pi}{6}$
- $\sin(2x) = \tan(x)$resposta$x = 0, \pi, \dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{7\pi}{4}$
- $\cot^{4}(x) = 4\csc^{2}(x) - 7$resposta$x = \dfrac{\pi}{6}, \dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{5\pi}{4}, \dfrac{7\pi}{4}, \dfrac{11\pi}{6}$
- $\cos(2x) + \csc^{2}(x) = 0$resposta$x = \dfrac{\pi}{2}, \dfrac{3\pi}{2}$
- $\tan^{3} \left( x \right) = 3\tan \left( x \right)$resposta$x = 0, \dfrac{\pi}{3}, \dfrac{2\pi}{3}, \pi, \dfrac{4\pi}{3}, \dfrac{5\pi}{3}$
- $\tan^{2} \left( x \right) = \dfrac{3}{2} \sec \left( x \right)$resposta$x = \dfrac{\pi}{3}, \dfrac{5\pi}{3}$
- $\cos^{3} \left( x \right) = -\cos \left( x \right)$resposta$x = \dfrac{\pi}{2}, \dfrac{3\pi}{2}$
- $\tan (2x) - 2\cos(x) = 0$resposta$x = \dfrac{\pi}{6}, \dfrac{\pi}{2}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}$
- $\csc^{3}(x) + \csc^{2}(x) = 4\csc(x) + 4$resposta$x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{3\pi}{2}, \dfrac{11\pi}{6}$
- $2\tan(x) = 1 - \tan^{2}(x)$resposta$x = \dfrac{\pi}{8}, \dfrac{5\pi}{8}, \dfrac{9\pi}{8}, \dfrac{13\pi}{8}$
- Sem solução
Nos Exercícios 43 a 58, resolva a equação, fornecendo as soluções exatas que estão no intervalo $[0, 2\pi)$.
- $x = 0, \dfrac{\pi}{7}, \dfrac{2\pi}{7}, \dfrac{3\pi}{7}, \dfrac{4\pi}{7}, \dfrac{5\pi}{7}, \dfrac{6\pi}{7}, \pi, \dfrac{8\pi}{7}, \dfrac{9\pi}{7}, \dfrac{10\pi}{7}, \dfrac{11\pi}{7}, \dfrac{12\pi}{7}, \dfrac{13\pi}{7}$
- $\sin(3x)\cos(x) = \cos(3x) \sin(x)$resposta$x=0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2}$
- $\cos(2x)\cos(x) + \sin(2x)\sin(x) = 1$resposta$x = 0$
- $\cos(5x)\cos(3x) - \sin(5x)\sin(3x) = \dfrac{\sqrt{3}}{2}$resposta$x = \dfrac{\pi}{48}, \dfrac{11\pi}{48}, \dfrac{13\pi}{48}, \dfrac{23\pi}{48}, \dfrac{25\pi}{48}, \dfrac{35\pi}{48}, \dfrac{37\pi}{48}, \dfrac{47\pi}{48}, \dfrac{49\pi}{48}, \dfrac{59\pi}{48}, \dfrac{61\pi}{48}, \dfrac{71\pi}{48}, \dfrac{73\pi}{48}, \dfrac{83\pi}{48}, \dfrac{85\pi}{48}, \dfrac{95\pi}{48}$
- $\sin(x) + \cos(x) = 1$resposta$x = 0, \dfrac{\pi}{2}$
- $\sin(x) + \sqrt{3} \cos(x) = 1$resposta$x = \dfrac{\pi}{2}, \dfrac{11\pi}{6}$
- $\sqrt{2} \cos(x) - \sqrt{2} \sin(x) = 1$resposta$x = \dfrac{\pi}{12}, \dfrac{17\pi}{12}$
- $\sqrt{3} \sin(2x) + \cos(2x) = 1$resposta$x= 0, \pi, \dfrac{\pi}{3}, \dfrac{4\pi}{3}$
- $\cos(2x) - \sqrt{3} \sin(2x) = \sqrt{2}$resposta$x = \dfrac{17 \pi}{24}, \dfrac{41 \pi}{24}, \dfrac{23\pi}{24}, \dfrac{47\pi}{24}$
- $3\sqrt{3}\sin(3x) - 3\cos(3x) = 3\sqrt{3}$resposta$x = \dfrac{\pi}{6}, \dfrac{5\pi}{18}, \dfrac{5\pi}{6}, \dfrac{17\pi}{18}, \dfrac{3\pi}{2}, \dfrac{29\pi}{18}$
- $\cos(3x) = \cos(5x)$resposta$x = 0, \dfrac{\pi}{4}, \dfrac{\pi}{2}, \dfrac{3\pi}{4}, \pi, \dfrac{5\pi}{4}, \dfrac{3\pi}{2}, \dfrac{7\pi}{4}$
- $\cos(4x) = \cos(2x)$resposta$x = 0, \dfrac{\pi}{3}, \dfrac{2\pi}{3}, \pi, \dfrac{4\pi}{3}, \dfrac{5\pi}{3}$
- $\sin(5x) = \sin(3x)$resposta$x = 0, \dfrac{\pi}{8}, \dfrac{3\pi}{8}, \dfrac{5\pi}{8}, \dfrac{7\pi}{8}, \pi, \dfrac{9\pi}{8}, \dfrac{11\pi}{8}, \dfrac{13\pi}{8}, \dfrac{15\pi}{8}$
- $\cos(5x) = -\cos(2x)$resposta$x = \dfrac{\pi}{7}, \dfrac{\pi}{3}, \dfrac{3\pi}{7}, \dfrac{5\pi}{7}, \pi, \dfrac{9\pi}{7}, \dfrac{11\pi}{7}, \dfrac{5\pi}{3}, \dfrac{13\pi}{7}$
- $\sin(6x) + \sin(x) = 0$resposta$x = \dfrac{2\pi}{7}, \dfrac{4\pi}{7}, \dfrac{6\pi}{7}, \dfrac{8\pi}{7}, \dfrac{10\pi}{7}, \dfrac{12\pi}{7}, \dfrac{\pi}{5}, \dfrac{3\pi}{5}, \pi, \dfrac{7\pi}{5}, \dfrac{9\pi}{5}$
- $x = \arcsin \left( \dfrac{-1 + \sqrt{5}}{2} \right) \approx 0.6662, \pi - \arcsin \left( \dfrac{-1 + \sqrt{5}}{2} \right) \approx 2.4754$
Nos Exercícios 59 a 70, resolva a desigualdade. Expressa a resposta exata em notação de intervalo, restringindo sua atenção a $0 \leq x \leq 2\pi$.
- $\left[ \pi, 2\pi \right]$
- $\tan \left( x \right) \geq \sqrt{3}$resposta$\left[ \dfrac{\pi}{3}, \dfrac{\pi}{2} \right) \cup \left[ \dfrac{4\pi}{3}, \dfrac{3\pi}{2} \right)$
- $\sec^{2} \left( x \right) \leq 4$resposta$\left[ 0, \dfrac{\pi}{3} \right] \cup \left[ \dfrac{2\pi}{3}, \dfrac{4\pi}{3} \right] \cup \left[ \dfrac{5\pi}{3}, 2\pi \right]$
- $\cos^{2} \left( x \right) > \dfrac{1}{2}$resposta$\left[ 0, \dfrac{\pi}{4} \right) \cup \left( \dfrac{3\pi}{4}, \dfrac{5\pi}{4} \right) \cup \left( \dfrac{7\pi}{4}, 2\pi \right]$
- $\cos \left( 2x \right) \leq 0$resposta$\left[ \dfrac{\pi}{4}, \dfrac{3\pi}{4} \right] \cup \left[ \dfrac{5\pi}{4}, \dfrac{7\pi}{4} \right]$
- $\sin \left( x + \dfrac{\pi}{3} \right) > \dfrac{1}{2}$resposta$\left[ 0, \dfrac{\pi}{2} \right) \cup \left( \dfrac{11\pi}{6}, 2\pi \right]$
- $\cot^{2} \left( x \right) \geq \dfrac{1}{3}$resposta$\left( 0, \dfrac{\pi}{3} \right] \cup \left[ \dfrac{2\pi}{3}, \pi \right) \cup \left( \pi, \dfrac{4\pi}{3} \right] \cup \left[ \dfrac{5\pi}{3}, 2\pi \right)$
- $2\cos(x) \geq 1$resposta$\left[0, \dfrac{\pi}{3}\right] \cup \left[\dfrac{5\pi}{3}, 2\pi\right]$
- $\sin(5x) \geq 5$respostaSem solução
- $\cos(3x) \leq 1$resposta$[0, 2\pi]$
- $\sec(x) \leq \sqrt{2}$resposta$\left[0, \dfrac{\pi}{4} \right] \cup \left(\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right) \cup \left[\dfrac{7\pi}{4}, 2\pi\right]$
- $\left[\text{arccot}(4), \pi \right) \cup \left[ \pi + \text{arccot}(4), 2\pi\right)$
Nos Exercícios 71 a 76, resolva a desigualdade. Expressa a resposta exata em notação de intervalo, restringindo sua atenção a $-\pi \leq x \leq \pi$.
- $\left( -\dfrac{\pi}{6}, \dfrac{\pi}{6} \right)$
- $\sin(x) > \dfrac{1}{3}$resposta$\left( \arcsin\left(\dfrac{1}{3}\right), \pi - \arcsin\left(\dfrac{1}{3}\right) \right)$
- $\sec \left( x \right) \leq 2$resposta$\left[ -\pi, -\dfrac{\pi}{2} \right) \cup \left[ -\dfrac{\pi}{3}, \dfrac{\pi}{3} \right] \cup \left( \dfrac{\pi}{2}, \pi \right]$
- $\sin^{2} \left( x \right) < \dfrac{3}{4}$resposta$\left( -\dfrac{2\pi}{3}, -\dfrac{\pi}{3} \right) \cup \left( \dfrac{\pi}{3}, \dfrac{2\pi}{3} \right)$
- $\cot \left( x \right) \geq -1$resposta$\left( -\pi, -\dfrac{\pi}{4} \right] \cup \left( 0, \dfrac{3\pi}{4} \right]$
- $\left[ -\dfrac{3\pi}{4}, \dfrac{\pi}{4} \right]$
Nos Exercícios 77 a 82, resolva a desigualdade. Expressa a resposta exata em notação de intervalo, restringindo sua atenção a $-2\pi \leq x \leq 2\pi$.
- $\left( -2\pi, -\dfrac{3\pi}{2} \right) \cup \left( -\dfrac{3\pi}{2}, -\pi \right) \cup \left( 0, \dfrac{\pi}{2} \right) \cup \left( \dfrac{\pi}{2}, \pi \right)$
- $\cos(x) \leq \dfrac{5}{3}$resposta$[-2\pi, 2\pi]$
- $\cot(x) \geq 5$resposta$\left(-2\pi, \text{arccot}(5) - 2\pi\right] \cup \left(-\pi, \text{arccot}(5) - \pi\right] \cup \left(0, \text{arccot}(5)\right] \cup \left(\pi, \pi + \text{arccot}(5)\right]$
- $\tan^{2} \left( x \right) \geq 1$resposta$\left[ -\dfrac{7\pi}{4}, -\dfrac{3\pi}{2} \right) \cup \left( -\dfrac{3\pi}{2}, -\dfrac{5\pi}{4} \right] \cup \left[ -\dfrac{3\pi}{4}, -\dfrac{\pi}{2} \right) \cup \left( -\dfrac{\pi}{2}, -\dfrac{\pi}{4} \right] \cup \left[ \dfrac{\pi}{4}, \dfrac{\pi}{2} \right) \cup \left( \dfrac{\pi}{2}, \dfrac{3\pi}{4} \right] \cup \left[ \dfrac{5\pi}{4}, \dfrac{3\pi}{2} \right) \cup \left( \dfrac{3\pi}{2}, \dfrac{7\pi}{4} \right]$
- $\sin(2x) \geq \sin(x)$resposta$\left[ -2\pi, -\dfrac{5\pi}{3} \right] \cup \left[ -\pi, -\dfrac{\pi}{3} \right] \cup \left[ 0, \dfrac{\pi}{3} \right] \cup \left[ \pi, \dfrac{5\pi}{3} \right]$
- $\left[ -\dfrac{11\pi}{6}, -\dfrac{7\pi}{6} \right] \cup \left[ \dfrac{\pi}{6}, \dfrac{5\pi}{6} \right] \cup \left\{ -\dfrac{\pi}{2}, \dfrac{3\pi}{2} \right\}$
Nos Exercícios 83 a 91, expresse o domínio da função usando a notação de intervalo estendida.
- $\displaystyle \bigcup_{k=-\infty}^{\infty} \left( 2k\pi, (2k+2)\pi \right)$
- $f(x) = \dfrac{\cos(x)}{\sin(x) + 1}$resposta$\displaystyle \bigcup_{k=-\infty}^{\infty} \left( \dfrac{(4k - 1)\pi}{2}, \dfrac{(4k + 3)\pi}{2} \right)$
- $f(x) = \sqrt{\tan^{2}(x) - 1}$resposta$\displaystyle \bigcup_{k=-\infty}^{\infty} \left\{ \left[ \dfrac{(4k + 1)\pi}{4}, \dfrac{(2k + 1)\pi}{2} \right) \cup \left( \dfrac{(2k + 1)\pi}{2}, \dfrac{(4k + 3)\pi}{4} \right] \right\}$
- $f(x) = \sqrt{2 - \sec(x)}$resposta$\displaystyle \bigcup_{k=-\infty}^{\infty} \left\{ \left[ \dfrac{(6k - 1)\pi}{3}, \dfrac{(6k + 1)\pi}{3} \right] \cup \left( \dfrac{(4k + 1)\pi}{2}, \dfrac{(4k + 3)\pi}{2} \right) \right\}$
- $f(x) = \csc(2x)$resposta$\displaystyle \bigcup_{k=-\infty}^{\infty} \left( \dfrac{k\pi}{2}, \dfrac{(k+1)\pi}{2} \right)$
- $f(x) = \dfrac{\sin(x)}{2 + \cos(x)}$resposta$(-\infty, \infty)$
- $f(x) = 3\csc(x) + 4\sec(x)$resposta$\displaystyle \bigcup_{k=-\infty}^{\infty} \left( \dfrac{k\pi}{2}, \dfrac{(k+1)\pi}{2} \right)$
- $f(x) = \ln\left( |\cos(x)| \right)$resposta$\displaystyle \bigcup_{k=-\infty}^{\infty} \left( \dfrac{(2k - 1)\pi}{2}, \dfrac{(2k+1)\pi}{2} \right)$
- $\displaystyle \bigcup_{k=-\infty}^{\infty} \left[ \dfrac{(4k - 1)\pi}{4}, \dfrac{(4k+1)\pi}{4} \right]$
- Com a ajuda de seus colegas, determine o número de soluções para $\sin(x) = \frac{1}{2}$ no intervalo $[0,2\pi)$. Em seguida, encontre o número de soluções para $\sin(2x) = \frac{1}{2}$, $\sin(3x) = \frac{1}{2}$ e $\sin(4x) = \frac{1}{2}$ no intervalo $[0,2\pi)$. Deverá surgir um padrão. Explique como esse padrão ajudaria a resolver equações como $\sin(11x) = \frac{1}{2}$. Agora, considere $\sin\left(\frac{x}{2}\right) = \frac{1}{2}$, $\sin\left(\frac{3x}{2}\right) = \frac{1}{2}$ e $\sin\left(\frac{5x}{2}\right) = \frac{1}{2}$. O que você encontra? Substitua $\dfrac{1}{2}$ por $-1$ e repita toda a exploração.