Giải thích các bước giải:

ĐKXĐ: $x\notin\{\dfrac{11}6\pi+k2\pi, \dfrac76\pi+k2\pi, \dfrac12\pi+k2\pi\}$

Ta có:
$\dfrac{(1-2\sin x)\cos x}{(1+2\sin x)(1-\sin x)}=\sqrt3$

$\to (1-2\sin x)\cos x=\sqrt3(1+2\sin x)(1-\sin x)$

$\to \cos x-2\sin x\cos x= \sqrt3(1+2\sin x-\sin x+2\sin^2x)$

$\to \cos x-2\sin x\cos x= \sqrt3(\sin x+1+2\sin^2x)$

$\to \cos x-\sin2x= \sqrt3(\sin x+\cos2x)$

$\to \cos x-\sqrt3\sin x=\sin2x+\sqrt3\cos2x$

$\to \dfrac12\cos x-\dfrac{\sqrt3}2\sin x=\dfrac12\sin2x+\dfrac{\sqrt3}{2}\cos2x$

$\to \cos(x+\dfrac13\pi)=\cos(2x-\dfrac16\pi)$

$\to x+\dfrac13\pi=2x-\dfrac16\pi+k2\pi\to x=\dfrac12\pi-2k\pi$

Hoặc $x+\dfrac13\pi=-(2x-\dfrac16\pi)+k2\pi\to x=-\dfrac1{18}\pi+\dfrac3k\pi$