Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{24} + k\dfrac{\pi}{2}\\x = \dfrac{5\pi}{12} + k\pi\end{array}\right.\quad (k\in \Bbb Z)$
Giải thích các bước giải:
$\cos x - \sqrt3\sin x = 2\sin3x$
$\Leftrightarrow \dfrac{1}{2}\cos x - \dfrac{\sqrt3}{2}\sin x = \sin3x$
$\Leftrightarrow \sin\left(\dfrac{\pi}{6} - x\right) = 3x$
$\Leftrightarrow \left[\begin{array}{l}\dfrac{\pi}{6} - x = 3x + k2\pi\\\dfrac{\pi}{6} - x = \pi - 3x + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}4x = \dfrac{\pi}{6} + k2\pi\\2x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{24} + k\dfrac{\pi}{2}\\x = \dfrac{5\pi}{12} + k\pi\end{array}\right.\quad (k\in \Bbb Z)$
