2căn 3 sin2x-cos3x=2-căn 3sin 2x cos3x mình cần gấp ạ

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:

$2\sqrt3\sin2x – \cos3x = 2- \sqrt3\sin2x\cos3x$

$\Leftrightarrow 2(\sqrt3\sin2x – 1) = \cos3x(1 – \sqrt3\sin2x)$

$\Leftrightarrow (\sqrt3\sin2x – 1)(\cos3x + 2) = 0$

$\Leftrightarrow \left[\begin{array}{l}\sin2x = \dfrac{1}{\sqrt3} \quad (nhận)\\\cos3x = -2\quad (loại)\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{1}{2}\arcsin\dfrac{1}{\sqrt3} + k\pi\\x = \dfrac{\pi}{2} – \dfrac{1}{2}\arcsin\dfrac{1}{\sqrt3} + k\pi\end{array}\right.\quad (k\in \Bbb Z)$