Question

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

2)cos^2(x) - sin2x =1  

Answer 1

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

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