2sin^2(x)+2sinxcos2x-1=0

Решите задачу:

$2sin^{2}x+2sinxcosx-1=0\\ \frac{2(1-cos2x)}{2}+2sinxcosx-1=0\\ 1-cos2x+2sinxcosx-1=0\\ 2sinxc0s2x-cos2x=0\\ cos2x(2sinx-1)=0\\ cos2x=0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2sinx-1=0\\ 2x=\frac{\pi}{2}+2pi*n\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \2sinx=1\ \ \ sinx=\frac{1}{2}\\ x=\frac{\pi}{4}+pi*n;\ x=\frac{\pi}{6}+2pi*n; \ x=\pi-\frac{\pi}{6}+2pi*n=\frac{5\pi}{6}+2pi*n$