$5.44 \operatorname{tg} 5 x-2 \operatorname{tg} 3 x=\operatorname{tg}^{2} 3 x \operatorname{tg} 5 x$.
5.44 Here $\cos 3 x \neq 0, \cos 5 x \neq 0$. Move $\operatorname{tg} 3 x$ to the right side of the equation:

$$
\begin{aligned}
& \operatorname{tg} 5 x-\operatorname{tg} 3 x=\operatorname{tg} 3 x+\operatorname{tg}^{2} 3 x \operatorname{tg} 5 x \\
& \operatorname{tg} 5 x-\operatorname{tg} 3 x=\operatorname{tg} 3 x(1+\operatorname{tg} 3 x \operatorname{tg} 5 x) \\
& \frac{\operatorname{tg} 5 x-\operatorname{tg} 3 x}{1+\operatorname{tg} 3 x \operatorname{tg} 5 x}=\operatorname{tg} 3 x
\end{aligned}
$$

(provided that $1+\operatorname{tg} 3 x \operatorname{tg} 5 x \neq 0$); $\operatorname{tg} 2 x=\operatorname{tg} 3 x ; 3 x=2 x+\pi k$; $x=\pi k$.

Answer: $x=\pi k, k \in Z$.