Помогите пожалуйста ) извините ,за почерк)

$\frac{\cos2x}{1-\sin 2x} =\cos x+\sin x \\ \frac{\cos 2x}{\sin^2x+\cos^2x-2\sin x\cos x} =\cos x+\sin x \\ \frac{(\cos x-\sin x)(\cos x+\sin x)}{(\cos x-\sin x)^2} =\cos x+\sin x \\ (\cos x+\sin x)-(\cos x+\sin x )(\cos x-\sin x)=0 \\ (\cos x+\sin x)(1-\cos x+\sin x)=0 \\$
cos x+ sin x=0 |: cos x
tg x=-1
x=-π/4+πn,n ∈ Z
$\cos x-\sin x=1$
Формула: $a\cos \pm b\sin x= \sqrt{a^2+b^2} \sin (x\pm \arcsin \frac{b}{ \sqrt{a^2+b^2} } )$

$\sqrt{2} \sin (x- \frac{\pi}{4} )=1 \\ \sin (x- \frac{\pi}{4} )= \frac{1}{ \sqrt{2}} \\ x-\frac{\pi}{4}=(-1)^k\cdot \frac{\pi}{4}+ \pi k,k \in Z \\ x=(-1)^k\cdot \frac{\pi}{4}+\frac{\pi}{4}+ \pi k,k \in Z$

$\sin ^3x+\cos ^3x=0|:\cos ^3x \\ tg^3x=-1 \\ tg x=-1 \\ x=-\frac{\pi}{4}+ \pi n,n \in Z$