$8.471 \frac{3 \tan x-\tan^{3} x}{2-\frac{1}{\cos ^{2} x}}=\frac{4+2 \cos \frac{6 x}{5}}{\cos 3 x+\cos x}$.
Solution. $\frac{\left(\frac{3 \sin x}{\cos x}-\frac{\sin ^{3} x}{\cos ^{3} x}\right) \cos ^{2} x}{2 \cos ^{2} x-1}=\frac{4+2 \cos \frac{6 x}{5}}{2 \cos 2 x \cdot \cos x} \Leftrightarrow$

$$
\begin{aligned}
& \Leftrightarrow \frac{\sin x \cdot\left(3 \cos ^{2} x-\sin ^{2} x\right)}{\cos x \cdot \cos 2 x}=\frac{2+\cos \frac{6 x}{5}}{\cos x \cdot \cos 2 x} \Leftrightarrow \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ \operatorname { cos } x \neq 0 , } \\
{ \operatorname { cos } 2 x \neq 0 , } \\
{ \operatorname { sin } x \cdot ( 3 - 4 \operatorname { sin } ^ { 2 } x ) = 2 + \operatorname { cos } \frac { 6 x } { 5 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\cos x \neq 0, \\
\cos 2 x \neq 0, \\
\sin 3 x-\cos \frac{6 x}{5}=2
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ \operatorname { cos } x \neq 0 , } \\
{ \operatorname { cos } 2 x \neq 0 , } \\
{ \operatorname { sin } 3 x = 1 , } \\
{ \operatorname { cos } \frac { 6 x } { 5 } = - 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x \neq \frac{\pi}{2}+\pi l, \\
x \neq \frac{\pi}{4}+\frac{\pi l}{2}, \\
x=\frac{\pi}{6}+\frac{2 \pi m}{3}, \\
x=\frac{5}{6}(\pi+2 \pi n)
\end{array}\right.\right.
\end{aligned}
$$

If $\frac{\pi}{6}+\frac{2 \pi m}{3}=\frac{5}{6}(\pi+2 \pi n)$, then $n=2 k, k \in Z$ (see the method of solving equation (4)). Therefore, the last system is equivalent to

$$
\left\{\begin{array}{l}
x \neq \frac{\pi}{2}+\pi l \\
x \neq \frac{\pi}{4}+\frac{\pi l}{2} \\
x=\frac{5}{6}(\pi+4 \pi k)
\end{array}\right.
$$

If $\frac{\pi}{2}+\pi l=\frac{5}{6}(\pi+4 \pi k)$, then $k=2+3 t, t \in Z ;$ the equation $\frac{\pi}{4}+\frac{\pi l}{2}=$ $=\frac{5}{6}(\pi+4 \pi k)$ has no solutions (see the method of solving equation (4)).

Answer: $\quad x=\frac{5}{6}(\pi+4 \pi k), k \neq 2+3 t$.