COS^4X+COS^4(X+Л/4)=1/4

$\cos^4x+\cos^4(x+ \frac{\pi}{4})= \frac{1}{4} \\ \\ \frac{(1+\cos2x)^2}{4}+\frac{(1-\sin2x)^2}{4} =\frac{1}{4}\\ \\ 1+2\cos2x+\cos^22x+1-2\sin2x+\sin^22x=1\\ 2\cos2x-2\sin2x=-2|:(-2)\\ \\ \sin2x -\cos2x=1\\ \\ \sqrt{2}\sin(2x-\frac{\pi}{4})=1\\ \\ 2x-\frac{\pi}{4}=(-1)^k\cdot\frac{\pi}{4}+\pi k,k \in Z\\ \\ 2x=(-1)^k\cdot \frac{\pi}{4}+\frac{\pi}{4}+\pi k,k \in Z\\ \\ x=(-1)^k\cdot \frac{\pi}{8}+\frac{\pi}{8}+\pi k,k \in Z$

Ответ: $(-1)^k\cdot \frac{\pi}{8}+\frac{\pi}{8}+\pi k,k \in Z$