Question

(Unifor - CE) No intervalo [-π , π], o número de soluções da equação 2.cos4x=1 é?

Answer 1

Olá!
  
    $2\cos(4x)=1\Leftrightarrow \cos(4x) = \dfrac{1}{2}.$

Lembre que os arcos cujo cosseno valem 1/2 são  $\dfrac{\pi}{3}+2k\pi,\;k\in\mathbb{Z}$  .
 
Ou seja, devemos ter

$4x = \dfrac{\pi}{3}+2k\pi\Leftrightarrow 4x = \dfrac{\pi + 6k\pi}{3}\Leftrightarrow x=\dfrac{\pi+6k\pi}{12}=\dfrac{\pi}{12}+\dfrac{k\pi}{2}\\ \\ \text{Agora,}\\ \\ -\pi\ \textless \ \dfrac{\pi}{12}+\dfrac{k\pi}{2}\ \textless \ \pi\Leftrightarrow -\pi\ \textless \ \dfrac{\pi+6k\pi}{12}\ \textless \ \pi \Leftrightarrow \\ \\ \\ \Leftrightarrow -12\pi\ \textless \ \pi+6k\pi\ \textless \ 12\pi\Leftrightarrow -12\pi\ \textless \ \pi(6k+1)\ \textless \ 12\pi\Leftrightarrow \\ \\ \\ \Leftrightarrow -12\ \textless \ 6k+1\ \textless \ 12\Leftrightarrow -13\ \textless \ 6k\ \textless \ 11\Leftrightarrow -\dfrac{13}{6}\ \textless \ k\ \textless \ \dfrac{11}{6}.$

Como  $k\in\mathbb{Z}$  ,  $-\dfrac{13}{6} = -2,1666..$  e  $\dfrac{11}{6} = 1,8333...$  então 

$-2\leqslant k\leqslant 1$,

ou seja, k pode assumir os valores -2, -1, 0 ou 1. Logo,

$x=\dfrac{\pi}{12}+\dfrac{k\pi}{2}\Rightarrow \left\{\begin{array}{lcr} x & = & \dfrac{\pi}{12}+\dfrac{(-2)\pi}{2}\\ \text{ou} & \; & \; \\ x & = & \dfrac{\pi}{12}+\dfrac{(-1)\pi}{2} \\ \text{ou} & \; & \; \\ x & = & \dfrac{\pi}{12}+ \dfrac{0\pi}{2}\\ \text{ou} & \; & \; \\ x & = & \dfrac{\pi}{12}+ \dfrac{1\pi}{2}\end{array}$

Logo, x pode assumir 4 valores no intervalo dado, e são eles:

$\left\{\dfrac{\pi}{12}-\pi, \dfrac{\pi}{12}-\dfrac{\pi}{2}, \dfrac{\pi}{12}, \dfrac{\pi}{12}+ \dfrac{\pi}{2}\right\} = \left\{-\dfrac{11\pi}{12}, -\dfrac{5\pi}{12}, \dfrac{\pi}{12}, \dfrac{7\pi}{12}\right\}.$




Bons estudos!