$5.20 \operatorname{tg} x+\operatorname{ctg} x=\frac{2}{\cos 4 x}$.

$5.20 \tan x + \cot x = \frac{2}{\cos 4x}$.
5.20 The equation is defined for $\sin x \neq 0, \cos x \neq 0, \cos 4 x \neq 0$.

$$
\text { Since } \operatorname{tg} x+\operatorname{ctg} x=\frac{2}{\sin 2 x} \text {, then } \frac{2}{\sin 2 x}=\frac{2}{\cos 4 x} \text {, from which }
$$

$$
\sin 2 x=\cos 4 x, \cos \left(\frac{\pi}{2}-2 x\right)-\cos 4 x=0
$$

$$
\sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{4}-3 x\right)=0
$$

## Therefore:

1) $\sin \left(\frac{\pi}{4}+x\right)=0 ; \frac{\pi}{4}+x=\pi k ; x_{1}=\frac{\pi}{4}(4 k-1)$;
2) $\sin \left(\frac{\pi}{4}-3 x\right)=0 ; \frac{\pi}{4}-3 x=-\pi k ; x_{2}=\frac{\pi}{12}(4 k+1)-$ this solution includes $x_{1}$, so $x=\frac{\pi}{12}(4 k+1)$.

Answer: $\frac{\pi}{12}(4 k+1), k \in Z$.