8.004. $\frac{\sin ^{2} 2 x-4 \sin ^{2} x}{\sin ^{2} 2 x+4 \sin ^{2} x-4}+1=2 \tan^{2} x$.
## Solution.

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

We have

$$
\begin{aligned}
& \frac{\sin ^{2} 2 x-4 \sin ^{2} x+\sin ^{2} 2 x+4 \sin ^{2} x-4}{\sin ^{2} 2 x+4 \sin ^{2} x-4}=2 \operatorname{tg}^{2} x \Leftrightarrow \\
& \Leftrightarrow \frac{2 \sin ^{2} 2 x-4}{\sin ^{2} 2 x+4 \sin ^{2} x-4}=2 \operatorname{tg}^{2} x, \frac{\sin ^{2} 2 x-2}{\sin ^{2} 2 x+4 \sin ^{2} x-4}=\operatorname{tg}^{2} x \Leftrightarrow
\end{aligned}
$$

$\Leftrightarrow \frac{\sin ^{2} 2 x-2}{\sin ^{2} 2 x+4 \sin ^{2} x-4}=\frac{1-\cos 2 x}{1+\cos 2 x} \Leftrightarrow$

$\Leftrightarrow \frac{1-\cos ^{2} 2 x-2}{1-\cos ^{2} 2 x+2-2 \cos 2 x-4}=\frac{1-\cos 2 x}{1+\cos 2 x} \Rightarrow \cos ^{3} 2 x+\cos ^{2} 2 x=0$, $\cos ^{2} 2 x(\cos 2 x+1)=0$.

## From this:

1) $\cos ^{2} 2 x=0, \quad \cos 2 x=0, \quad 2 x=\frac{\pi}{2}+\pi k, \quad k \in Z$,

$$
x_{1}=\frac{\pi}{4}+\frac{\pi k}{2}=\frac{\pi}{4}(2 k+1) \quad k \in Z
$$

2) $\cos 2 x+1=0, \cos 2 x=-1, \quad 2 x=\pi+2 \pi n, \quad n \in Z$,

$$
x_{2}=\frac{\pi}{2}+\pi n=\frac{\pi}{2}(2 n+1), \quad n \in Z
$$

Combining the solutions, we get $x=\frac{\pi}{4}+\frac{\pi m}{2}=\frac{\pi}{4}(2 m+1), m \in Z$.

Answer: $\quad x=\frac{\pi}{4}(2 m+1) \quad m \in Z$.