Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{4}+ k\dfrac{\pi}{2}\\x = -\dfrac{\pi}{2} + k\pi\end{array}\right. \quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\cos3x = -\cos x$
$\Leftrightarrow \cos3x = \cos(\pi - x)$
$\Leftrightarrow \left[\begin{array}{l}3x = \pi - x + k2\pi\\3x = x -\pi + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{4}+ k\dfrac{\pi}{2}\\x = -\dfrac{\pi}{2} + k\pi\end{array}\right. \quad (k \in \Bbb Z)$
