8.121. $\frac{\sin ^{2} x-2}{\sin ^{2} x-4 \cos ^{2} \frac{x}{2}}=\tan ^{2} \frac{x}{2}$.
## Solution.

DZ: $\left\{\begin{array}{l}\cos ^{2} \frac{x}{2} \neq 0, \\ \sin ^{2} \frac{x}{2}-4 \cos ^{2} \frac{x}{2} \neq 0 .\end{array}\right.$

Rewrite the given equation as

$$
\begin{aligned}
& \frac{\sin ^{2} x-2}{\sin ^{2} x-2(1+\cos x)}=\frac{1-\cos x}{1+\cos x}, \frac{1-\cos ^{2} x-2}{1-\cos ^{2} x-2-2 \cos x}-\frac{1-\cos x}{1+\cos x}=0 \\
& \frac{\cos ^{2} x+1}{\cos ^{2} x+2 \cos x+1}-\frac{1-\cos x}{1+\cos x}=0, \frac{\cos ^{2} x+1}{(1+\cos x)^{2}}-\frac{1-\cos x}{1+\cos x}=0 \Rightarrow \\
& \Rightarrow \cos ^{2} x+1-(1-\cos x)(1+\cos x)=0 \Leftrightarrow 2 \cos ^{2} x=0, \quad \cos x=0 \\
& x=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2 k+1), k \in Z
\end{aligned}
$$

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