3. The equation is equivalent to

$$
\begin{gathered}
5(\operatorname{tg} x+\operatorname{ctg} 3 x)=\operatorname{tg} 2 x-\operatorname{tg} x \Leftrightarrow 5 \frac{\cos 2 x}{\cos x \sin 3 x}=\frac{\sin x}{\cos x \cos 2 x} \Leftrightarrow \\
\left\{\begin{array} { l } 
{ 1 0 \operatorname { c o s } ^ { 2 } 2 x = \operatorname { c o s } 2 x - \operatorname { c o s } 4 x , } \\
{ \operatorname { c o s } x \neq 0 , \operatorname { c o s } 2 x \neq 0 , \operatorname { s i n } 3 x \neq 0 . }
\end{array} \Leftrightarrow \left[\begin{array}{l}
\cos 2 x=-\frac{1}{4} \\
\cos 2 x=\frac{1}{3}
\end{array}\right.\right.
\end{gathered}
$$
Answer: $\pm \frac{1}{2} \arccos \left(-\frac{1}{4}\right)+\pi n, \pm \frac{1}{2} \arccos \left(\frac{1}{3}\right)+\pi n, n \in Z$.

Answer to variant $6-2: \pm \frac{1}{2} \arccos \left(-\frac{1}{2}\right)+\pi n, \pm \frac{1}{2} \arccos \left(\frac{1}{3}\right)+\pi n, n \in Z$.