Sinx^4(x)+cos^4(x)=cos^2(2x)+1/4 Нужно подробно

Решите задачу:

$sin^4(x)+cos^4(x)=cos^2(2x)+\frac{1}{4}\\\\ sin^4(x)+2*sin^2(x)*cos^2(x)+cos^4(x)-2*sin^2(x)*cos^2(x)=\\=cos^2(2x)+\frac{1}{4}\\\\ \ \ [sin^2(x)+cos^2(x)]^2-2*sin^2(x)*cos^2(x)=cos^2(2x)+\frac{1}{4}\\\\ 1^2-\frac{1}{2}*[2*sin(x)*cos(x)]^2=cos^2(2x)+\frac{1}{4}\\\\ \frac{3}{4}-\frac{1}{2}*[sin(2x)]^2=cos^2(2x)\\\\ 3-2*sin^2(2x)=4cos^2(2x)\\\\ 3-2*[1-cos^2(2x)]=4cos^2(2x)\\\\ 3-2+2cos^2(2x)=4cos^2(2x)\\\\ 1=2cos^2(2x)\\\\ 1=2*\frac{1+cos(4x)}{2}\\\\ 1=1+cos(4x)\\\\ cos(4x)=0\\\\$

$4x=\frac{\pi}{2}+\pi n,\ n\in Z\\\\ x=\frac{\pi}{8}+\frac{\pi n}{4},\ n\in$