8.016. $\operatorname{ctg}^{4} 2 z+\sin ^{-4} 2 z=25$.

8.016. $\cot^{4} 2 z+\sin ^{-4} 2 z=25$.
## Solution.

Domain of definition: $\sin 2 z \neq 0$.

From the condition

$$
\begin{aligned}
& \frac{\cos ^{4} 2 z}{\sin ^{4} 2 z}+\frac{1}{\sin ^{4} 2 z}-25=0 \Leftrightarrow \cos ^{4} 2 z+1-25 \sin ^{4} 2 z=0 \Leftrightarrow \\
& \Leftrightarrow\left(\cos ^{2} 2 z\right)^{2}+1-25 \sin ^{4} 2 z=0, \quad\left(1-\sin ^{2} 2 z\right)^{2}+1-25 \sin ^{4} 2 z=0 \Leftrightarrow \\
& \Leftrightarrow 12 \sin ^{4} 2 z+\sin ^{2} 2 z-1=0
\end{aligned}
$$

Solving the equation as a biquadratic equation in $\sin 2 z$, we get $\sin 2 z= \pm \frac{1}{2}$, from which $2 z= \pm \frac{\pi}{6}+\pi k, z= \pm \frac{\pi}{12}+\frac{\pi k}{2}=\frac{\pi}{12}(6 k \pm 1), k \in Z$.

Answer: $\quad z=\frac{\pi}{12}(6 k \pm 1), k \in Z$.