8.362. $\sin ^{6} x+\cos ^{6} x=\frac{7}{16}$.

8.362. $\sin ^{6} x+\cos ^{6} x=\frac{7}{16}$.
## Solution.

Let's write the equation as

$$
\begin{aligned}
& \left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x\right)-\frac{7}{16}=0 \Leftrightarrow \\
& \Leftrightarrow\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-3 \sin ^{2} x \cos ^{2} x-\frac{7}{16}=0 \Leftrightarrow \\
& \Leftrightarrow 1-3 \sin ^{2} x \cos ^{2} x-\frac{7}{16}=0 \Leftrightarrow 16-12\left(4 \sin ^{2} x \cos ^{2} x\right)-7=0 \Leftrightarrow
\end{aligned}
$$

$$
\begin{aligned}
& \Leftrightarrow \sin ^{2} 2 x=\frac{3}{4}, \sin 2 x= \pm \frac{\sqrt{3}}{2}, 2 x= \pm \frac{\pi}{3}+\pi k \\
& x= \pm \frac{\pi}{6}+\frac{\pi k}{2}=\frac{\pi}{6}(3 k \pm 1), k \in Z
\end{aligned}
$$

Answer: $x=\frac{\pi}{6}(3 k \pm 1), k \in Z$.