8.277. $\operatorname{ctg} x-\operatorname{tg} x-2 \operatorname{tg} 2 x-4 \operatorname{tg} 4 x+8=0$.

8.277. $\cot x - \tan x - 2 \tan 2x - 4 \tan 4x + 8 = 0$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}\sin x \neq 0, \\ \cos x \neq 0, \\ \cos 2 x \neq 0, \\ \cos 4 x \neq 0 .\end{array}\right.$

Rewrite the equation as

$$
\begin{aligned}
& \left(\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}\right)-2 \tan 2 x-4 \tan 4 x+8=0 \Leftrightarrow \\
& \Leftrightarrow \frac{\cos ^{2} x-\sin ^{2} x}{\sin x \cos x}-2 \tan 2 x-4 \tan 4 x+8=0 \Leftrightarrow \\
& \Leftrightarrow \frac{2 \cos 2 x}{\sin 2 x}-2 \tan 2 x-4 \tan 4 x+8=0 \Leftrightarrow \\
& \Leftrightarrow 2 \cdot\left(\frac{\cos 2 x}{\sin 2 x}-\frac{\sin 2 x}{\cos 2 x}\right)-4 \tan 4 x+8=0 \Leftrightarrow \\
& \Leftrightarrow \frac{2\left(\cos ^{2} 2 x-\sin ^{2} 2 x\right)}{\sin 2 x \cos 2 x}-4 \tan 4 x+8=0 \Leftrightarrow \\
& \Leftrightarrow \frac{4 \cos 4 x}{\sin 4 x}-4 \tan 4 x+8=0 \Leftrightarrow \frac{\cos 4 x}{\sin 4 x}-\frac{\sin 4 x}{\cos 4 x}+2=0 \Leftrightarrow \\
& \Leftrightarrow \frac{\cos ^{2} 4 x-\sin ^{2} 4 x}{\sin 4 x \cos 4 x}+2=0 \Leftrightarrow \frac{2 \cos 8 x}{\sin 8 x}+2=0 \Leftrightarrow \cot 8 x+1=0 \Leftrightarrow \\
& \Leftrightarrow \cot 8 x=-1, 8 x=\frac{3 \pi}{4}+\pi k, x=\frac{3 \pi}{32}+\frac{\pi k}{8}=\frac{\pi}{32}(4 k+3), k \in \mathbb{Z}
\end{aligned}
$$

Answer: $x=\frac{\pi}{32}(4 k+3), k \in \mathbb{Z}$.