Алгебра: Sin^2*3x/4 - корень2/2=sinx-cos^2*3x/4+1

Решите задачу:

$sin^2\frac{3x}{4}-\frac{\sqrt2}{2}=sinx-cos^2 \frac{3x}{4}+1\\\\\underbrace {sin^2 \frac{3x}{4}+ cos^2\frac{3x}{4}}_{1}-\frac{\sqrt2}{2}-1=sinx\\\\sinx=-\frac{\sqrt2}{2}\\\\x=(-1)^{n}\cdot arcsin(-\frac{\sqrt2}{2})+\pi n,\; n\in Z\\\\x=(-1)^{n}\cdot (-\frac{\pi}{4})+\pi n=(-1)^{n+1}\cdot \frac{\pi}{4}+\pi n,\; n\in Z$