8.296. $\cos ^{-4} z=64 \cos ^{2} 2 z$.
Solution.

Domain of definition: $\cos z \neq 0$.

From the condition we have

$\frac{1}{\cos ^{4} z}=(8 \cos 2 z)^{2} \Leftrightarrow\left[\begin{array}{l}\frac{1}{\cos ^{2} z}=8 \cos 2 z, \cos 2 z>0, \\ \frac{1}{\cos ^{2} z}=-8 \cos 2 z, \cos 2 z<0,\end{array} \Leftrightarrow\right.$

$\Leftrightarrow\left[\begin{array}{l}\frac{2}{1+\cos 2 z}=8 \cos 2 z, \\ \frac{1}{1+\cos 2 z}=-8 \cos 2 z,\end{array} \Leftrightarrow\left[\begin{array}{l}4 \cos ^{2} 2 z+4 \cos 2 z-1=0, \\ 4 \cos ^{2} 2 z+4 \cos 2 z+1=0,\end{array} \Leftrightarrow\right.\right.$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0562.jpg?height=309&width=1040&top_left_y=526&top_left_x=154)

1) $2 \cos ^{2} z-1=\frac{-1+\sqrt{2}}{2}, \cos z= \pm \frac{\sqrt{1+\sqrt{2}}}{2}, z_{1}= \pm \arccos \frac{\sqrt{1+\sqrt{2}}}{2}+\pi k, k \in Z$;
2) $2 z= \pm \frac{2}{3} \pi+2 \pi n, z_{2}= \pm \frac{\pi}{3}+\pi n=\frac{\pi}{3}(3 n \pm 1), n \in Z$.

Answer: $z_{1}= \pm \arccos \frac{\sqrt{1+\sqrt{2}}}{2}+\pi k, k \in Z ; z_{2}=\frac{\pi}{3}(3 n \pm 1), n \in Z$.