2. Solve the equation $\cos x \cos 2x \cos 4x \cos 8x = \frac{1}{16}$.
Solution. We can write the equation in the form

$16 \cos x \cos 2 x \cos 4 x \cos 8 x=1$, or $(2 \cos x) \cdot(2 \cos 2 x) \cdot(2 \cos 4 x) \cdot(2 \cos 8 x)=1$.

If we multiply the last equation by $\sin x$, we get

$$
(2 \cos x \sin x)(2 \cos 2 x)(2 \cos 4 x)(2 \cos 8 x)=\sin x
$$

and if we use the identity $\sin 2 \alpha=2 \sin \alpha \cos \alpha$, we get

$$
\begin{aligned}
& (2 \sin 2 x \cos 2 x)(2 \cos 4 x)(2 \cos 8 x)=\sin x \\
& (2 \sin 4 x \cos 4 x)(2 \cos 8 x)=\sin x \\
& 2 \sin 8 x \cos 8 x=\sin x \\
& \sin 16 x=\sin x
\end{aligned}
$$

If we use the identity $\sin \alpha-\sin \beta=2 \sin \frac{\alpha-\beta}{2} \cos \frac{\alpha+\beta}{2}$ for the equation $\sin 16 x-\sin x=0$, we get $2 \sin \frac{15 x}{2} \cos \frac{17 x}{2}=0$. The equation $\sin \frac{15 x}{2}=0$ has solutions $\frac{15 x}{2}=k \pi, k \in \mathbb{Z}$, i.e., $x=\frac{2 k \pi}{15}, k \in \mathbb{Z}$. The equation $\cos \frac{17 x}{2}=0$ has solutions $\frac{17 x}{2}=\frac{\pi}{2}+k \pi, k \in \mathbb{Z}$, i.e., $x=\frac{\pi+2 k \pi}{17}, k \in \mathbb{Z}$.

The solutions to the equation are $x=\frac{2 k \pi}{15}, k \in \mathbb{Z}$ and $x=\frac{\pi+2 k \pi}{17}, k \in \mathbb{Z}$, where $k \neq 15 s$ and $2 k+1 \neq 17 s, s \in \mathbb{Z}$.