Đáp án:
$\left[ \begin{array}{l}x = \dfrac{\pi}{2} + k2\pi\\x = \pm \dfrac{\pi}{6} + k2\pi\end{array} \right. \,\,\,\,(k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l} \sin2x = 2\cos x+\sqrt3(\sin x - 1)\\ \Leftrightarrow 2\sin x\cos x - 2\cos x - \sqrt3(\sin x - 1)=0\\ \Leftrightarrow 2\cos x(\sin x - 1) -\sqrt3(\sin x -1)=0\\ \Leftrightarrow (\sin x -1)(2\cos x - \sqrt3) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x = 1\\\cos x = \dfrac{\sqrt3}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi}{2} + k2\pi\\x = \pm \dfrac{\pi}{6} + k2\pi\end{array} \right. \,\,\,\,(k \in \Bbb Z) \end{array}$
