8.326. $(\cos x-\sin x)^{2}+\cos ^{4} x-\sin ^{4} x=0.5 \sin 4 x$.
## Solution.

From the condition we get

$$
\begin{aligned}
& \cos ^{2} x-2 \cos x \sin x+\sin ^{2} x+\left(\cos ^{2} x+\sin ^{2} x\right)\left(\cos ^{2} x-\sin ^{2} x\right)- \\
& -0.5 \sin 4 x=0 \Leftrightarrow 1-\sin 2 x+\cos 2 x-\sin 2 x \cos 2 x=0 \Leftrightarrow \\
& \Leftrightarrow(1+\cos 2 x)-(\sin 2 x+\sin 2 x \cos 2 x)=0 \Leftrightarrow \\
& \Leftrightarrow(1+\cos 2 x)-\sin 2 x(1+\cos 2 x)=0 \Leftrightarrow(1+\cos 2 x)(1-\sin 2 x)=0
\end{aligned}
$$

From this

1) $1+\cos 2 x=0, \cos 2 x=-1,2 x_{1}=\pi+2 \pi n, x_{1}=\frac{\pi}{2}+\pi n=\frac{\pi}{2}(2 n+1), n \in Z$.
2) $1-\sin 2 x=0, \sin 2 x=1,2 x_{2}=\frac{\pi}{2}+2 \pi k, x_{2}=\frac{\pi}{4}+\pi k=\frac{\pi}{4}(4 k+1), k \in Z$.

Answer: $x_{1}=\frac{\pi}{2}(2 n+1) ; x_{2}=\frac{\pi}{4}(4 k+1)$, where $n$ and $k \in Z$.