$2\sin 2x-2\cos2x=\frac{\cos x+\cos3x}{\cos x-\sin x}$ Solve the equation :
$$2\sin 2x-2\cos2x=\frac{\cos x+\cos3x}{\cos x-\sin x}$$
my try :
$$\cos x-\sin x \neq 0 \to \cos x \neq \sin x$$
$$x\neq k\pi+\frac{\pi}{4}$$
$$2(\sin 2x -\cos 2x)=\frac{\cos x+\cos3x}{\cos x-\sin x}$$
$$2(2\sin x \cos x -\cos 2x)=\frac{\cos x+\cos3x}{\cos x-\sin x}$$
$$2(2\sin x \cos x -(1-2\sin^2x)=\frac{\cos x+\cos3x}{\cos x-\sin x}$$
then ?

Hint:
$\cos3x+\cos x=2\cos2x\cos x$
$\sin2x-\cos2x=\cos x(\cos x+\sin x)$
$\iff \cos^2x+\cos2x=\sin x\cos x$
Divide both sides by $\cos^2x$
