8.215. $\left(\cos ^{-2} 2 x+\operatorname{tg}^{2} 2 x\right)\left(\sin ^{-2} 2 x+\operatorname{ctg}^{2} 2 x\right)=4 \sin ^{-2} 4 x+5$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}\cos 2 x \neq 0, \\ \sin 2 x \neq 0 .\end{array}\right.$

Write the equation in the form:

$$
\left(\frac{1}{\cos ^{2} 2 x}+\frac{\sin ^{2} 2 x}{\cos ^{2} 2 x}\right) \cdot\left(\frac{1}{\sin ^{2} 2 x}+\frac{\cos ^{2} 2 x}{\sin ^{2} 2 x}\right)=\frac{4}{\sin ^{2} 4 x}+5 \Leftrightarrow
$$

$$
\begin{aligned}
& \Leftrightarrow \frac{1+\sin ^{2} 2 x}{\cos ^{2} 2 x} \cdot \frac{1+\cos ^{2} 2 x}{\sin ^{2} 2 x}-\frac{4}{\sin ^{2} 4 x}-5=0 \Leftrightarrow \\
& \Leftrightarrow \frac{1+\cos ^{2} 2 x+\sin ^{2} 2 x+\sin ^{2} 2 x \cos ^{2} 2 x}{\sin ^{2} 2 x \cos ^{2} 2 x}-\frac{4}{\sin ^{2} 4 x}-5=0 \Leftrightarrow \\
& \Leftrightarrow \frac{2+\sin ^{2} 2 x \cos ^{2} 2 x}{\sin ^{2} 2 x \cos ^{2} 2 x}-\frac{4}{\sin ^{2} 4 x}-5=0 \Leftrightarrow \\
& \Leftrightarrow \frac{2}{\sin ^{2} 2 x \cos ^{2} 2 x}+1-\frac{4}{\sin ^{2} 4 x}-5=0 \Leftrightarrow \\
& \Leftrightarrow \frac{2}{\sin ^{2} 2 x \cos ^{2} 2 x}-\frac{4}{\sin ^{2} 4 x}-4=0 \Leftrightarrow \frac{8}{\sin ^{2} 4 x}-\frac{4}{\sin ^{2} 4 x}-4=0 \Leftrightarrow \\
& \Leftrightarrow \frac{4}{\sin ^{2} 4 x}=4 \Leftrightarrow \sin ^{2} 4 x=1, \sin 4 x= \pm 1,4 x=\frac{\pi}{2}+\pi k, \\
& x=\frac{\pi}{8}+\frac{\pi k}{4}=\frac{\pi}{8}(2 k+1), k \in Z
\end{aligned}
$$

$$
\text { Answer: } x=\frac{\pi}{8}(2 k+1), k \in Z
$$