8.184. $\cos z \cos 2z \cos 4z \cos 8z=\frac{1}{16}$
## Solution.

Multiplying both sides of the equation by $16 \sin z \neq 0$, we get

$8(2 \sin z \cos z) \cos 2 z \cos 4 z \cos 8 z=\sin z \Leftrightarrow 8 \sin 2 z \cos 2 z \cos 4 z \cos 8 z=\sin z$, $4(2 \sin 2 z \cos 2 z) \cos 4 z \cos 8 z=\sin z, 4 \sin 4 z \cos 4 z \cos 8 z=\sin z$, $2(2 \sin 4 z \cos 4 z) \cos 8 z=\sin z, 2 \sin 8 z \cos 8 z=\sin z$,

$$
\sin 16 z-\sin z=0 \Leftrightarrow 2 \cos \frac{16 z+z}{2} \sin \frac{16 z-z}{2}=0, \cos \frac{17 z}{2} \sin \frac{15 z}{2}=0
$$

From this, either $\cos \frac{17 z}{2}=0, \frac{17 z}{2}=\frac{\pi}{2}+\pi k, z_{1}=\frac{\pi}{17}+\frac{2}{17} \pi k=\frac{\pi}{17}(2 k+1)$, $k \in Z$, or $\sin \frac{15 z}{2}=0, \frac{15 z}{2}=\pi k, z_{2}=\frac{2}{15} \pi k, k \in Z$

From the set of values $z_{2}=\frac{2}{15} \pi k, k \in Z$, we need to exclude those values of $z$ for which $\sin z=0$, i.e., $z=\pi n, n \in Z$, which occur when $k=15 n$. Similarly, from the set of values $z_{1}=\frac{\pi}{17}(2 k+1), k \in Z$, we exclude the values $z=\pi n, n \in Z$, which occur when $2 k+1=17$; $2 k+1=51$; $2 k+1=85$; ..., or $k=8$; $k=25$; $k=42$; ..., i.e., $k=8+17 l$, $l \in Z$. Thus, for the original equation, we find the following set of solutions:

$$
z_{1}=\frac{2 \pi k}{15}, k \neq 15 l, k \in Z, l \in Z
$$

or

$$
z_{2}=\frac{\pi}{17}(2 k+1), k \neq 17 l+8, k \in Z, l \in Z
$$

Answer: $z_{1}=\frac{2 \pi k}{15}, k \neq 15 l, z_{2}=\frac{\pi}{17}(2 k+1), k \neq 17 l+8, k \in Z, l \in Z$