Solve the equation

$$
2 \cos 5 x+2 \cos 4 x+2 \cos 3 x+2 \cos 2 x+2 \cos x+1=0
$$
I. solution. Seeking the solution in the interval $[0 ; 2 \pi)$, it is clear that $x \neq 0$, so $\frac{x}{2} \neq 0, \pi$, which means $\sin \frac{x}{2} \neq 0$. Therefore, the equation is equivalent to

$$
\begin{gathered}
2 \cos 5 x \sin \frac{x}{2}+2 \cos 4 x \sin \frac{x}{2}+2 \cos 3 x \sin \frac{x}{2}+2 \cos 2 x \sin \frac{x}{2}+ \\
+2 \cos x \sin \frac{x}{2}+\sin \frac{x}{2}=0 \quad(x \neq 0)
\end{gathered}
$$

Let's transform the left side of the equation using the identity (which follows from the addition formula of the sine function) valid for any integer $k$:

$$
\sin \left(k+\frac{1}{2}\right) \alpha-\sin \left(k-\frac{1}{2}\right) \alpha=2 \cos k \alpha \sin \frac{\alpha}{2}
$$

with $\alpha=x$ and $k=1,2,3,4,5$:

$$
\begin{aligned}
& \quad\left(\sin \left(5+\frac{1}{2}\right) x-\sin \left(5-\frac{1}{2}\right) x\right)+\left(\sin \left(4+\frac{1}{2}\right) x-\sin \left(4-\frac{1}{2}\right) x\right)+ \\
& +\left(\sin \left(3+\frac{1}{2}\right) x-\sin \left(3-\frac{1}{2}\right) x\right)+\left(\sin \left(2+\frac{1}{2}\right) x-\sin \left(2-\frac{1}{2}\right) x\right)+ \\
& +\left(\sin \left(1+\frac{1}{2}\right) x-\sin \left(1-\frac{1}{2}\right) x\right)+\sin \frac{x}{2}=\sin \frac{11}{2} x=0
\end{aligned}
$$

We seek the non-zero solutions of the equation $\sin \frac{11}{2} x=0$ in the interval $[0 ; 2 \pi)$: $\frac{11}{2} x=\pi t, x=\frac{2 t \pi}{11}=\frac{2 \pi}{11}, \frac{4 \pi}{11}, \frac{6 \pi}{11}, \frac{8 \pi}{11}, \frac{10 \pi}{11}$, $\frac{12 \pi}{11}, \frac{14 \pi}{11}, \frac{16 \pi}{11}, \frac{18 \pi}{11}, \frac{20 \pi}{11}$.

II. solution. For the complex number $\varepsilon=\cos x+i \sin x$, the $k$-th power of $\varepsilon$ for any integer $k$ is $\varepsilon^{k}=\cos k x+i \sin k x$ according to the de Moivre formula; furthermore, $(\cos k x+i \sin k x)+(\cos (-k x)+i \sin (-k x))=2 \cos k x$. The left side of our equation is therefore the real part of the sum

$$
S=\varepsilon^{-5}+\varepsilon^{-4}+\varepsilon^{-3}+\varepsilon^{-2}+\varepsilon^{-1}+1+\varepsilon+\varepsilon^{2}+\varepsilon^{3}+\varepsilon^{4}+\varepsilon^{5}
$$

which is $S$ itself, since the imaginary part of $S$ is zero due to the oddness of the sine function. According to the formula for the sum of a geometric series:

$$
S=\frac{\varepsilon^{6}-\varepsilon^{-5}}{\varepsilon-1}
$$

thus the equation of the problem is equivalent to $\varepsilon^{6}=\varepsilon^{-5}, \varepsilon \neq 1$, or $\varepsilon^{11}=1 \neq \varepsilon$. The solutions of this equation are:

$$
\varepsilon=\cos \frac{2 t \pi}{11}+i \sin \frac{2 t \pi}{11}, \quad \text { i.e. } \quad x=\frac{2 t \pi}{11}, \quad t=1,2, \ldots, 10
$$

III. solution. First, we show that $x=\frac{2 \pi}{11}$ is a solution to the equation. Consider a regular 11-sided polygon in the plane with a Cartesian coordinate system, centered at the origin, with vertices $A_{0}, A_{1}, \ldots, A_{10}$ in positive order, where $A_{0}$ is the point $(1 ; 0)$. Then the first coordinate of $A_{k}$ is exactly $\cos \frac{2 k \pi}{11}$, so the sum of the first coordinates of the points $A_{i}$ is exactly

$$
2 \cos 5 x+2 \cos 4 x+2 \cos 3 x+2 \cos 2 x+2 \cos x+1
$$

The rotation by $\frac{2 \pi}{11}$ around $O$ maps each vector $\overrightarrow{O A_{i}}$ to the next one, so the sum of the vectors is 0, and therefore the sum of the first coordinates of the points $A_{i}$ is also 0. The same reasoning shows that for every $1 \leq k \leq 10$, $x=\frac{2 k \pi}{11}$ is a solution to the equation. Of course, adding any integer multiple of $2 \pi$ to each solution gives new solutions, so for every integer $k$ not divisible by 11, $x=\frac{2 k \pi}{11}$ is a solution to the equation. We show that there are no other solutions.

By repeatedly applying the addition formulas, we get:

$$
\begin{gathered}
\cos 2 x=2 \cos ^{2} x-1, \quad \cos 3 x=4 \cos ^{3} x-3 \cos x, \\
\cos 4 x=8 \cos ^{4} x-8 \cos ^{2} x+1
\end{gathered}
$$

and

$$
\cos 5 x=16 \cos ^{5} x-20 \cos ^{3} x+5 \cos x
$$

Therefore, $x$ is a solution to the equation if and only if $a=\cos x$ is a root of the equation

$$
32 a^{5}+16 a^{4}-32 a^{3}-12 a^{2}+4 a+1=0
$$

According to the above, the numbers

$$
a_{1}=\cos \frac{2 \pi}{11}, \quad a_{2}=\cos \frac{4 \pi}{11}, \quad a_{3}=\cos \frac{6 \pi}{11}, \quad a_{4}=\cos \frac{8 \pi}{11}, \quad a_{5}=\cos \frac{10 \pi}{11}
$$

satisfy the equation. Since these are five distinct real numbers, the fifth-degree equation cannot have other solutions. Since the angles listed above are exactly those whose cosine is one of the $a_{i}$, we have indeed given all the solutions to the original equation.