ĐK: $\sin x+\cos x\ne0 \Leftrightarrow \tan x\ne-1 \Leftrightarrow x\ne\dfrac{-\pi}{4}+k\pi,k\in\mathbb{Z}$
Phương trình đã cho tương đương với:
$(1-\sin^2x)(\cos x-1)=2(\cos x+\sin x)(1+\sin x)$
$\Leftrightarrow (1+\sin x)(1-\sin x)(\cos x-1)=2(\sin x+\cos x)(1+\sin x)$
$\Leftrightarrow (1+\sin x)(1+\sin x+\cos x+\sin x\cos x)=0$
$\Leftrightarrow (1+\sin x)^2(1+\cos x)=0$
$\Leftrightarrow \left[\begin{array}{l}\sin x=-1\\\cos x=-1\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x=-\dfrac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{array}\right.,k\in\mathbb{Z}$