$8.4864 \cos ^{2} 2 x-\operatorname{tg} 4 x=\operatorname{ctg} 2 x$.

$8.4864 \cos ^{2} 2 x-\tan 4 x=\cot 2 x$.
Solution.

$4 \cos ^{2} 2 x=\operatorname{ctg} 2 x+\operatorname{tg} 4 x \Leftrightarrow \frac{\cos 2 x \cdot \cos 4 x+\sin 2 x \cdot \sin 4 x}{\cos 4 x \cdot \sin 2 x}=4 \cos ^{2} 2 x \Leftrightarrow$

$\Leftrightarrow 4 \cos ^{2} 2 x=\frac{\cos 2 x}{\cos 4 x \cdot \sin 2 x} \Leftrightarrow\left\{\begin{array}{l}\cos 2 x \cdot(4 \cos 2 x \cdot \sin 2 x \cdot \cos 4 x-1)=0, \\ \cos 4 x \neq 0, \\ \sin 2 x \neq 0\end{array} \Leftrightarrow\right.$ $\Leftrightarrow\left\{\begin{array}{l}(1-\sin 8 x) \cos 2 x=0 \\ \cos 4 x \neq 0, \\ \sin 2 x \neq 0\end{array}\right.$

1) $\left\{\begin{array}{l}\cos 2 x=0, \\ \cos 4 x \neq 0, \\ \sin 2 x \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}\cos 2 x=0, \\ \cos ^{2} 2 x \neq \sin ^{2} 2 x\end{array} \Leftrightarrow \cos 2 x=0 \Leftrightarrow x=\frac{\pi}{4}+\frac{\pi k}{2}\right.\right.$.
2) $\sin 8 x=1 \Leftrightarrow x=\frac{\pi}{16}+\frac{\pi k}{4}$.

Answer: $x_{1}=\frac{\pi}{4}(2 k+1), x_{2}=\frac{\pi}{16}(4 k+1)$.