8.102. $\sin ^{4} x+\cos ^{4} x=\sin 2 x-0.5$.
Solution.

We have

$$
\begin{aligned}
& \left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x-\sin 2 x+0.5=0 \Leftrightarrow 1-\frac{1}{2} \sin ^{2} 2 x- \\
& -\sin 2 x+\frac{1}{2}=0, \sin ^{2} 2 x+2 \sin 2 x-3=0
\end{aligned}
$$

Solving this equation as a quadratic equation in terms of $\sin 2 x$, we get $\sin 2 x=-3, \varnothing$, or $\sin 2 x=1, 2 x=\frac{\pi}{2}+2 \pi n, x=\frac{\pi}{4}+\pi n=\frac{\pi}{4}(4 n+1), \quad n \in Z$.

Answer: $x=\frac{\pi}{4}(4 n+1), n \in Z$.