8.108. $2 \sin ^{2} z+\tan ^{2} z=2$.
## Solution.

Domain of definition: $\cos z \neq 0$.

From the condition

$$
\begin{aligned}
& 2 \sin ^{2} z+\frac{\sin ^{2} z}{\cos ^{2} z}-2=0 \Leftrightarrow 2 \sin ^{2} z+\frac{\sin ^{2} z}{1-\sin ^{2} z}-2=0 \Rightarrow \\
& \Rightarrow 2 \sin ^{2} z\left(1-\sin ^{2} z\right)+\sin ^{2} z-2\left(1-\sin ^{2} z\right)=0, \quad 2 \sin ^{4} z-5 \sin ^{2} z+2=0
\end{aligned}
$$

Solving this equation as a biquadratic equation in terms of $\sin z$, we get $\sin z= \pm \frac{\sqrt{2}}{2}, z=\frac{\pi}{4}+\frac{\pi k}{2}=\frac{\pi}{4}(2 k+1), k \in Z$.

Answer: $\quad z=\frac{\pi}{4}(2 k+1), k \in Z$.