Solve the following system of equations:

$$
\begin{aligned}
& \tan x_{i}+3 \cot x_{i}=2 \tan x_{i+1} \quad(i=1,2, \ldots, n-1) \\
& \tan x_{n}+3 \cot x_{n}=2 \tan x_{1} .
\end{aligned}
$$
Let's denote $\left(\operatorname{tg} x_{i}\right)$ by $y_{i}$, with this our system of equations transforms as follows:

$$
\begin{aligned}
& y_{i}+\frac{3}{y_{i}}=2 y_{i+1} \quad(i=1,2, \ldots, n-1) \\
& y_{n}+\frac{3}{y_{n}}=2 y_{1}
\end{aligned}
$$

If this has a solution, then $y_{i} \neq 0$, and if $y_{1}, y_{2}, \ldots, y_{n}$ is a solution, then $-y_{1},-y_{2}, \ldots,-y_{n}$ is also a solution. Therefore, we can assume that $y_{1}>0$ and then by (2) $y_{i}>0$ for all $i$. Based on the arithmetic and geometric mean inequality, for the left side of (2),

$$
y_{i+1}=\frac{y_{i}+\frac{3}{y_{i}}}{2} \geq \sqrt{3}
$$

thus $y_{i} \geq \sqrt{3}$, which means $3 / y_{i} \leq \sqrt{3}$ for all $1 \leq i \leq n$. Now let's add up the $n$ equations under (2). After rearranging,

$$
\frac{3}{y_{1}}+\frac{3}{y_{2}} \ldots+\frac{3}{y_{n}}=y_{1}+y_{2}+\ldots+y_{n}
$$

The value on the left side is at most $n \sqrt{3}$, and the value on the right side is at least $n \sqrt{3}$, and equality can only hold if $y_{i}=\sqrt{3}$ for all $i$.

Therefore, (2) has two solutions: $y_{1}=y_{2}=\ldots=y_{n}=\sqrt{3}$, and $y_{1}=y_{2}=\ldots=y_{n}=-\sqrt{3}$. The solutions to the original system of equations are thus all $n$-tuples of numbers $x_{1}, x_{2}, \ldots, x_{n}$ such that $\operatorname{tg} x_{1}=\operatorname{tg} x_{2}=\ldots=\operatorname{tg} x_{n}=\sqrt{3}$, i.e., $x_{i}=60^{\circ}+k_{i} \cdot 180^{\circ}\left(k_{i}\right.$ integer $)$, or $\operatorname{tg} x_{1}=\operatorname{tg} x_{2}=\ldots=\operatorname{tg} x_{n}=-\sqrt{3}$, i.e., $x_{i}=-60^{\circ}+l_{i} \cdot 180^{\circ}\left(l_{i}\right.$ integer).

Remark. The formula $y_{i+1}=\left(y_{i}+3 / y_{i}\right) / 2$ is the so-called Newton iteration formula for determining $\sqrt{3}$. Starting from any positive $y_{1}$ value, the values of $y_{2}, y_{3}$, etc., increasingly approximate $\sqrt{3}$, and starting from a negative $y_{1}$ value, they increasingly approximate $-\sqrt{3}$. Therefore, two equal terms can only appear in the sequence if all terms are equal to $\sqrt{3}$ (or $-\sqrt{3}$).