1)sin^2(x) + sin2x=1 2)cos^2(x) - sin2x =1

1.
$\sin^2x + \sin2x=1$
$\sin^2x + 2\sin x\cos x=\sin^2x +\cos^2x$
$2\sin x\cos x=\cos^2x$
$2\sin x\cos x-\cos^2x=0$
$\cos x(2\sin x-\cos x)=0$
$\left[\begin{array}{l} \cos x=0 \\ 2\sin x-\cos x=0 \end{array}$
$\left[\begin{array}{l} x= \frac{ \pi }{2}+ \pi n \\ 2\mathrm{tg} x-1=0 \end{array}$
$\left[\begin{array}{l} x= \frac{ \pi }{2}+ \pi n \\ \mathrm{tg} x= \frac{1}{2} \end{array}$
$\left[\begin{array}{l} x= \frac{ \pi }{2}+ \pi n \\ x= \mathrm{arctg} \frac{1}{2} + \pi n \end{array}, \ n\in Z$

2.
$\cos^2x - \sin2x =1$
$\cos^2x -2 \sin x\cos x =\sin^2x+\cos^2x$
$-2 \sin x\cos x =\sin^2x$
$\sin^2x+2 \sin x\cos x=0$
$\sin x(\sin x+2 \cos x)=0$
$\left[\begin{array}{l} \sin x=0 \\ \sin x+2 \cos x=0 \end{array}$
$\left[\begin{array}{l} x= \pi n \\ \mathrm{tg}x+2=0 \end{array}$
$\left[\begin{array}{l} x= \pi n \\ \mathrm{tg}x=-2 \end{array}$
$\left[\begin{array}{l} x= \pi n \\ x=-\mathrm{arctg}2+ \pi n \end{array}, \ n\in Z$