ĐK: $\sin x\neq 0\Leftrightarrow x\neq k\pi,k\in Z$
$+\sin(2x-\frac{3\pi}{2})=\sin(2x+\frac{\pi}{2}-2\pi)=\sin(2x+\frac{\pi}{2})=\cos2x$
$pt\Leftrightarrow (\cos x-\sin x)(1-\sqrt{3}\cos4x)=2\cos2x\sin x$$\Leftrightarrow (\cos x-\sin x)(1-\sqrt{3}\cos4x)=2\sin x(\cos x-\sin x)(\cos x+\sin x)$
$\Leftrightarrow (\cos x-\sin x)(1-\sqrt{3}\cos4x-\sin2x-2\sin^2x)=0$
$\Leftrightarrow (\cos x-\sin x)(\cos2x-\sin2x-\sqrt{3}(\cos2x-\sin2x)(\cos2x+\sin2x))=0$
$\Leftrightarrow (\cos x-\sin x)\left[(\cos2x-\sin2x)(1-\sqrt{3}(\cos2x+\sin2x) {} \right]=0$
$\Leftrightarrow (\cos x-\sin x)(\cos2x-\sin2x)(1-\sqrt{6}\sin(x+\frac{\pi}{4}))=0$
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