Đáp án:+$[_{x = - \frac{\pi }{{12}} + k\pi }^{x = \frac{\pi }{{12}} + k\pi },k \in Z$
+$[_{x = - \frac{{5\pi }}{{12}} + k\pi }^{x = \frac{{5\pi }}{{12}} + k\pi },k \in Z$
Giải thích các bước giải:
$\begin{array}{l}
4{\cos ^2}2x - 3 = 0\\
< = > {\cos ^2}2x = \frac{3}{4}\\
< = > [_{\cos 2x = \frac{{ - \sqrt 3 }}{2}}^{\cos 2x = \frac{{\sqrt 3 }}{2}}\\
+ \cos 2x = \frac{{\sqrt 3 }}{2} = \cos \frac{\pi }{6}\\
< = > [_{2x = - \frac{\pi }{6} + k2\pi }^{2x = \frac{\pi }{6} + k2\pi }\\
< = > [_{x = - \frac{\pi }{{12}} + k\pi }^{x = \frac{\pi }{{12}} + k\pi },k \in Z\\
+ \cos 2x = - \frac{{\sqrt 3 }}{2} = \cos \frac{{5\pi }}{6}\\
< = > [_{2x = - \frac{{5\pi }}{6} + k2\pi }^{2x = \frac{{5\pi }}{6} + k2\pi }\\
< = > [_{x = - \frac{{5\pi }}{{12}} + k\pi }^{x = \frac{{5\pi }}{{12}} + k\pi },k \in Z
\end{array}$
