Đáp án:
$\left[\begin{array}{l}x = -\dfrac{\pi}{12} +k\pi\\x =- \dfrac{5\pi}{6}+ k2\pi\end{array}\right.\quad (k\in \Bbb Z)$
Giải thích các bước giải:
$\sqrt3\cos\left(x + \dfrac{\pi}{2}\right) + \sin\left(x - \dfrac{\pi}{2}\right) = 2\sin2x$
$\Leftrightarrow -\sqrt3\sin x - \cos x = 2\sin2x$
$\Leftrightarrow \dfrac{\sqrt3}{2}\sin x + \dfrac{1}{2}\cos x = \sin (-2x)$
$\Leftrightarrow \sin\left(x + \dfrac{\pi}{6}\right) = \sin(-2x)$
$\Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{6} = - x + k2\pi\\x + \dfrac{\pi}{6} = \pi + 2x + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{12} +k\pi\\x =- \dfrac{5\pi}{6}+ k2\pi\end{array}\right.\quad (k\in \Bbb Z)$
