$\cos x -\sin x =\sin2x - 2\cos^2x$

$\Leftrightarrow \cos x - \sin x = 2\sin x\cos x - 2\cos^2x$

$\Leftrightarrow \cos x - \sin x = 2\cos x(\sin x - \cos x)$

$\Leftrightarrow \cos x -\sin x - 2\cos x(\sin x -\cos x)= 0$

$\Leftrightarrow (\cos x -\sin x)(1 + 2\cos x)= 0$

$\Leftrightarrow \sqrt2\cos\left(x +\dfrac{\pi}{4}\right)(1+2\cos x)= 0$

$\Leftrightarrow \left[\begin{array}{l}\cos\left(x +\dfrac{\pi}{4}\right)= 0\\\cos x = -\dfrac12\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x +\dfrac{\pi}{4}=\dfrac{\pi}{2} + k\pi\\x = \pm \dfrac{2\pi}{3} + k2\pi\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{4}+ k\pi\\x = \pm \dfrac{2\pi}{3} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$

Vậy phương trình có tập nghiệm

$S =\left\{\pm \dfrac{2\pi}{3} + k2\pi,\ \dfrac{\pi}{4} + k\pi\ \Bigg|\ k\in\Bbb Z\right\}$