8.426 \frac{1-2 \cos ^{2} x}{\sin x \cdot \cos x}+2 \tan 2 x+\cot ^{3} 4 x=3
$$
Solution. Transform the original equation:

$$
\begin{aligned}
& -\frac{2 \cos 2 x}{\sin 2 x}+\frac{2}{\operatorname{ctg} 2 x}+\operatorname{ctg}^{3} 4 x=3 \Leftrightarrow \frac{2\left(1-\operatorname{ctg}^{2} 2 x\right)}{\operatorname{ctg} 2 x}+\operatorname{ctg}^{3} 4 x=3 \Leftrightarrow \\
& \Leftrightarrow-4 \operatorname{ctg} 4 x+\operatorname{ctg}^{3} 4 x=3 \Leftrightarrow(1+\operatorname{ctg} 4 x) \cdot\left(\operatorname{ctg}^{2} 4 x-\operatorname{ctg} 4 x-3\right)=0 \Leftrightarrow
\end{aligned}
$$

$$
\Leftrightarrow\left[\begin{array} { l } 
{ \operatorname { c t g } 4 x = - 1 , } \\
{ \operatorname { c t g } 4 x = \frac { 1 \pm \sqrt { 1 3 } } { 2 } }
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=\frac{3 \pi}{16}+\frac{\pi k}{4} \\
x=\frac{1}{4} \operatorname{arcctg} \frac{1 \pm \sqrt{13}}{2}+\frac{\pi k}{4}
\end{array}\right.\right.
$$

Answer: $x_{1}=\frac{\pi}{16}(4 k+3), x_{2}=\frac{1}{4}\left(\operatorname{arcctg} \frac{1 \pm \sqrt{13}}{2}+\pi k\right)$.