Question

(2senx - √3)( 2cosx - √2)= 0 ???????????

Answer 1

Resolver a equação trigonométrica

$(2\,\mathrm{sen\,}x-\sqrt{3})(2\cos x-\sqrt{2})=0$

Um produto é zero somente se algum dos fatores é zero. Logo, devemos ter

$\begin{array}{rcl} 2\,\mathrm{sen\,}x-\sqrt{3}=0&\quad\textsf{ou}\quad&2\cos x-\sqrt{2}=0\\\\ 2\,\mathrm{sen\,}x=\sqrt{3}&\quad\textsf{ou}\quad&2\cos x=\sqrt{2}\\\\ \mathrm{sen\,}x=\dfrac{\sqrt{3}}{2}&\quad\textsf{ou}\quad&\cos x=\dfrac{\sqrt{2}}{2}\\\\ \mathrm{sen\,}x=\mathrm{sen\,}\dfrac{\pi}{3}&\quad\textsf{ou}\quad&\cos x=\cos \dfrac{\pi}{4} \end{array}$

__________

Resolva as equações acima separadamente, e depois faça a união das soluções:

• $\mathrm{sen\,}x=\mathrm{sen\,}\dfrac{\pi}{3}$

$\begin{array}{rcl}x=\dfrac{\pi}{3}+k\cdot 2\pi &\quad\textsf{ou}\quad&x=\pi-\dfrac{\pi}{3}+k\cdot 2\pi\\\\ x=\dfrac{\pi}{3}+k\cdot 2\pi &\quad\textsf{ou}\quad&x=\dfrac{3\pi-\pi}{3}+k\cdot 2\pi\\\\ x=\dfrac{\pi}{3}+k\cdot 2\pi &\quad\textsf{ou}\quad&x=\dfrac{2\pi}{3}+k\cdot 2\pi \end{array}$

com $k$ inteiro.

• $\cos x=\cos \dfrac{\pi}{4}$

$x=\pm\,\dfrac{\pi}{4}+k\cdot 2\pi\\\\\\ \begin{array}{rcl}x=\dfrac{\pi}{4}+k\cdot 2\pi &\quad\textsf{ou}\quad&x=-\,\dfrac{\pi}{4}+k\cdot 2\pi \end{array}$

com $k$ inteiro.

__________

Portanto o conjunto solução é

$S=\left\{x\in \mathbb{R}:~~x=\theta+k\cdot 2\pi,~~\theta\in \Big\{\!-\dfrac{\pi}{4},\,\dfrac{\pi}{4},\,\dfrac{\pi}{3},\,\dfrac{2\pi}{3}\Big\},~~k\in \mathbb{Z}\right\}.$

Bons estudos! :-)