1-sin2x = sinx-cosx ​

1-sin2x = sinx-cosx

sin²x+cos²x-2·sin x·cos x = sin x - cos x

(sin x - cos x)² = sin x - cos x

(sin x - cos x)(sin x - cos x - 1) = 0

1) sin x - cos x = 0

tg x = 1

$x_1=\frac{\pi}{4} +\pi k,\ k\in Z$

2) sin x - cos x - 1 = 0

sin x - cos x = 1

$\frac{\sqrt{2} }{2}\sin x- \frac{\sqrt{2} }{2}\cos x=\frac{\sqrt{2} }{2}\\ -\sqrt{2} \cos(\frac{\pi}{4}+x)=\frac{\sqrt{2} }{2}\\\cos(\frac{\pi}{4}+x)=-\frac{1}{2} \\\\\frac{\pi}{4}+x=\б\frac{2\pi}{3}+2\pi n\\ \\ x=-\frac{\pi}{4}\б\frac{2\pi}{3}+2\pi n\\ \\ x_{2}=-\frac{11\pi}{12}+2\pi n;\ x_{3}=\frac{5\pi}{12}+2\pi n; \ n \in Z$

Ответ: $\frac{\pi}{4}+\pi k;\ -\frac{11\pi}{12}+2\pi n;\ \frac{5\pi}{12}+2\pi n; \ k,n \in Z$

1-sin2x = sinx-cosx ​

1-sin2x = sinx-cosx