Đáp án:
$\left[\begin{array}{l}x = \dfrac{1}{2}\arcsin\dfrac{1}{\sqrt3} + k\pi\\x = \dfrac{\pi}{2} - \dfrac{1}{2}\arcsin\dfrac{1}{\sqrt3} + k\pi\end{array}\right.\quad (k\in \Bbb Z)$
Giải thích các bước giải:
$2\sqrt3\sin2x - \cos3x = 2- \sqrt3\sin2x\cos3x$
$\Leftrightarrow 2(\sqrt3\sin2x - 1) = \cos3x(1 - \sqrt3\sin2x)$
$\Leftrightarrow (\sqrt3\sin2x - 1)(\cos3x + 2) = 0$
$\Leftrightarrow \left[\begin{array}{l}\sin2x = \dfrac{1}{\sqrt3} \quad (nhận)\\\cos3x = -2\quad (loại)\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{1}{2}\arcsin\dfrac{1}{\sqrt3} + k\pi\\x = \dfrac{\pi}{2} - \dfrac{1}{2}\arcsin\dfrac{1}{\sqrt3} + k\pi\end{array}\right.\quad (k\in \Bbb Z)$
