8.270. $\sin ^{3} x \cos 3 x+\cos ^{3} x \sin 3 x+0.375=0$.
## Solution.

We have $\sin ^{2} x(2 \sin x \cos 3 x)+\cos ^{2} x(2 \cos x \sin 3 x)+2 \cdot 0.375=0$. Using the formula $\sin \alpha \cos \beta=\frac{1}{2}(\sin (\alpha-\beta)+\sin (\alpha+\beta))$, we find

$$
\begin{aligned}
& \sin ^{2} x(-\sin 2 x+\sin 4 x)+\cos ^{2} x(\sin 2 x+\sin 4 x)+0.75=0 \Leftrightarrow \\
& \Leftrightarrow-\sin ^{2} x \sin 2 x+\sin ^{2} x \sin 4 x+\cos ^{2} x \sin 2 x+\cos ^{2} x \sin 4 x+0.75=0 \Leftrightarrow \\
& \Leftrightarrow\left(-\sin ^{2} x \sin 2 x+\cos ^{2} x \sin 2 x\right)+\left(\sin ^{2} x \sin 4 x+\cos ^{2} x \sin 4 x\right)+0.75=0 \Leftrightarrow \\
& \Leftrightarrow \sin 2 x\left(\cos ^{2} x-\sin ^{2} x\right)+\sin 4 x\left(\sin ^{2} x+\cos ^{2} x\right)+0.75=0 \Leftrightarrow \\
& \Leftrightarrow \sin 4 x+2 \sin 4 x+1.5=0,3 \sin 4 x=-1.5, \sin 4 x=-\frac{1}{2},
\end{aligned}
$$

$$
4 x=(-1)^{k+1} \frac{\pi}{6}+\pi k, x=(-1)^{k+1} \frac{\pi}{24}+\frac{\pi k}{4}, k \in Z
$$

Answer: $x=(-1)^{k+1} \frac{\pi}{24}+\frac{\pi k}{4}, k \in Z$.