8.195. $\frac{1-\sin ^{6} z-\cos ^{6} z}{1-\sin ^{4} z-\cos ^{4} z}=2 \cos ^{2} 3 z$.
## Solution.

Let's write the equation as

$$
\begin{aligned}
& \frac{1-\left(\left(\sin ^{2} z\right)^{3}+\left(\cos ^{2} z\right)^{3}\right)}{1-\left(\sin ^{4} z+\cos ^{4} z\right)}=2 \cos ^{2} 3 z \Leftrightarrow \\
& \Leftrightarrow \frac{1-\left(\sin ^{2} z+\cos ^{2} z\right)\left(\sin ^{4} z-\sin ^{2} z \cos ^{2} z+\cos ^{4} z\right)}{1-\left(\left(\sin ^{2} z+\cos ^{2} z\right)^{2}-2 \sin ^{2} z \cos ^{2} z\right)}=2 \cos ^{2} 3 z \Leftrightarrow \\
& \Leftrightarrow \frac{1-1\left(\left(\sin ^{2} z+\cos ^{2} z\right)^{2}-3 \sin ^{2} z \cos ^{2} z\right)}{1-1+2 \sin ^{2} z \cos ^{2} z}=2 \cos ^{2} 3 z \Leftrightarrow \\
& \Leftrightarrow \frac{1-1+3 \sin ^{2} z \cos ^{2} z}{2 \sin ^{2} z \cos ^{2} z}=2 \cos ^{2} 3 z, 2 \cos ^{2} 3 z=\frac{3}{4}, \cos 3 z= \pm \frac{\sqrt{3}}{2}
\end{aligned}
$$

$3 z= \pm \frac{\pi}{6}+\pi k, z= \pm \frac{\pi}{18}+\frac{\pi k}{3}=\frac{\pi}{18}(6 k \pm 1), k \in \mathbb{Z}$.

Answer: $z=\frac{\pi}{18}(6 k \pm 1), k \in \mathbb{Z}$.