Điều kiện: $\left\{\begin{array}{l}\sin x\ne-1\\\cos x\ne-1\end{array}\right.$
Phương trình đã cho tương đương với: $\dfrac{\sin x(1-\cos^2x)}{1+\cos x}+\dfrac{\cos x(1-\sin^2x)}{1+\sin x}=\cos2x+2\cos x-1$
$\Leftrightarrow \sin x(1-\cos x)+\cos x(1-\sin x)=2\cos x-2\sin^2x$
$\Leftrightarrow (\sin x-\cos x)(1+2\sin x)=0$
$\Leftrightarrow \left[\begin{array}{l}\sin x=\cos x\\\sin x=-\dfrac{1}{2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x=\dfrac{\pi}{4}+k\pi\\x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{array}\right. ,k\in\mathbb{Z}$.