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Điều kiện: $\left\{ \begin{array}{l} \sin x>0\\ \sin x\ne\frac{1}{4}\\ \cos x\ne0 \end{array} \right.$
Ta có:
$2\sqrt{3\sin x}=\frac{3\tan x}{2\sqrt{\sin x}-1}-\sqrt3$
$\Leftrightarrow 4\sqrt3\sin x-2\sqrt{3\sin x}=3\tan x-2\sqrt{3\sin x}+\sqrt3$
$\Leftrightarrow
4\sin x=\sqrt3\tan x+1$
$\Leftrightarrow 2\sin2x=\sqrt3\sin x+\cos x$
$\Leftrightarrow \sin 2x=\frac{\sqrt3}{2}\sin x+\frac{1}{2}\cos x$
$\Leftrightarrow \sin 2x=\sin\Big(x+\frac{\pi}{6}\Big)$
$\Leftrightarrow \left[ \begin{array}{l} 2x=
x+\frac{\pi}{6}+2k\pi\\2x=\frac{5\pi}{6}-x+2k\pi \end{array} \right. (k\in\mathbb{Z})$
$\Leftrightarrow
\left[ \begin{array}{l} x= \frac{\pi}{6}+2k\pi\\x=\frac{5\pi}{18}+\frac{2k}{3}\pi \end{array} \right. (k\in\mathbb{Z})$
Kết hợp với điều kiện: $x\in\{
\frac{\pi}{6}+2k\pi ,\frac{5\pi}{18}+2k\pi,\frac{17\pi}{18}+2k\pi | k\in\mathbb{Z}\}$
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