$5.18 \sin ^{4} x+\cos ^{4} x=\cos ^{2} 2 x+0.25$.
5.18 Let's use the identity $1=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}$, from which $\sin ^{4} x+\cos ^{4} x=1-2 \sin ^{2} x \cos ^{2} x=1-\frac{1}{2} \sin ^{2} 2 x$.

## Then we get

$1-\frac{1}{2} \sin ^{2} 2 x=1-\sin ^{2} 2 x+\frac{1}{4} ; 1-2 \sin ^{2} 2 x=0 ;$

$\cos 4 x=0, 4 x=\frac{\pi}{2}+\pi k ; x=\frac{\pi}{8}(2 k+1)$.

Answer: $\quad x=\frac{\pi}{8}(2 k+1), k \in Z$.