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Phương trình tương đương với:
$3-4\sin^22x=2\cos2x(1+2\sin x)$
$\Leftrightarrow 1+2(1-2\sin^22x)-2\cos 2x-4\cos 2x\sin x=0$
$\Leftrightarrow 1+2(\cos4x-\cos2x)-2(\sin3x-\sin x)=0$
$\Leftrightarrow 1-4\sin3x\sin x-2\sin 3x+2\sin x=0$
$\Leftrightarrow (1+2\sin x)(1-2\sin3x)=0$
$\Leftrightarrow \left[ \begin{array}{l} \sin x=\frac{-1}{2}\\\sin3x=\frac{1}{2} \end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l} x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\\x=\frac{\pi}{18}+k\frac{2\pi}{3}\\x=\frac{5\pi}{18}+k\frac{2\pi}{3} \end{array} \right.,k\in\mathbb{Z}$
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