$5.46 \operatorname{ctg} x-\operatorname{tg} x-2 \operatorname{tg} 2 x-4 \operatorname{tg} 4 x+8=0$.
5.46 Noting that $\operatorname{ctg} x-\operatorname{tg} x=\frac{1-\operatorname{tg}^{2} x}{\operatorname{tg} x}=2 \operatorname{ctg} 2 x$, we transform the left side of the equation:

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

From here $\operatorname{ctg} 8 x=-1,8 x=\frac{3 \pi}{4}+\pi k ; x=\frac{\pi}{32}(4 k+3)$.

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