8.220. $\frac{1}{\operatorname{tg}^{2} 2 x+\cos ^{-2} 2 x}+\frac{1}{\operatorname{ctg}^{2} 2 x+\sin ^{-2} 2 x}=\frac{2}{3}$.

8.220. $\frac{1}{\tan^{2} 2 x+\cos ^{-2} 2 x}+\frac{1}{\cot^{2} 2 x+\sin ^{-2} 2 x}=\frac{2}{3}$.
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:

$$
\frac{1}{\frac{\sin ^{2} 2 x}{\cos ^{2} 2 x}+\frac{1}{\cos ^{2} 2 x}}+\frac{1}{\frac{\cos ^{2} 2 x}{\sin ^{2} 2 x}+\frac{1}{\sin ^{2} 2 x}}=\frac{2}{3} \Leftrightarrow
$$

$\Leftrightarrow \frac{\cos ^{2} 2 x}{\sin ^{2} 2 x+1}+\frac{\sin ^{2} 2 x}{\cos ^{2} 2 x+1}-\frac{2}{3}=0 \Leftrightarrow$

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

$\Leftrightarrow \frac{1+\cos 4 x}{3-\cos 4 x}+\frac{1-\cos 4 x}{3+\cos 4 x}-\frac{2}{3}=0 \Leftrightarrow$

$\Leftrightarrow 3(1+\cos 4 x)(3+\cos 4 x)+3(3-\cos 4 x)(1-\cos 4 x)-$

$-2(3-\cos 4 x)(3+\cos 4 x)=0 \Leftrightarrow \cos ^{2} 4 x=0, \cos 4 x=0,4 x=\frac{\pi}{2}+\pi k$,

$x=\frac{\pi}{8}+\frac{\pi k}{4}=\frac{\pi}{8}(2 k+1), k \in Z$

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