Solve the system of equations

$$
\begin{aligned}
& \cos x+\cos y+\cos z=\frac{3 \sqrt{3}}{2} \\
& \sin x+\sin y+\sin z=\frac{3}{2}
\end{aligned}
$$
I. solution. We square both equations and then add the resulting equations. Taking into account the identity $\cos ^{2} \alpha+\sin ^{2} \alpha=1$, we obtain:

$$
\begin{gathered}
3+2(\cos x \cos y+\cos x \cos z+\cos y \cos z)+ \\
+2(\sin x \sin y+\sin x \sin z+\sin y \sin z)=\frac{27}{4}+\frac{9}{4}=\frac{36}{4}=9
\end{gathered}
$$

from which:

$$
(\cos x \cos y+\sin x \sin y)+(\cos x \cos z+\sin x \sin z)+(\cos y \cos z+\sin y \sin z)=3
$$

Applying the addition formula $\cos \alpha \cos \beta+\sin \alpha \sin \beta=\cos (\alpha-\beta)$ to all three terms: $\cos (x-y)+\cos (x-z)+$ $\cos (y-z)=3$. Since the value of the cosine function cannot be greater than 1, this equation can only be satisfied if $\cos (x-y)=\cos (x-z)=\cos (y-z)=1$, that is, $(x-y),(x-z),(y-z)$ are integer multiples of $2 \pi$:

$$
y=x+2 k_{2} \pi ; \quad z=x+2 k_{3} \pi
$$

where $k_{2}, k_{3}$ are arbitrary integers. Then $\cos x=\cos y=\cos z$ and $\sin x=\sin y=\sin z$; substituting these into the original equations:

$$
\cos x+\cos x+\cos x=\frac{3 \sqrt{3}}{2}, \quad \sin x+\sin x+\sin x=\frac{3}{2}
$$

in other words

$$
\cos x=\frac{\sqrt{3}}{2}, \quad \sin x=\frac{1}{2}
$$

from which:

$$
x=\frac{\pi}{6}+2 k_{1} \pi
$$

where $k_{1}$ is an arbitrary integer. Therefore, the solution is:

$$
x=\frac{\pi}{6}+2 k_{1} \pi, \quad y=\frac{\pi}{6}+2 k_{2} \pi, \quad z=\frac{\pi}{6}+2 k_{3} \pi
$$

where $k_{1}, k_{2}, k_{3}$ are arbitrary integers.

II. solution. Imagine the problem as if we had to represent the vector $\frac{3 \sqrt{3}}{2}+i \frac{3}{2}$ as the sum of three unit vectors in the complex plane. By introducing the unit-length complex unknowns $s=\cos x+i \sin x, t=\cos y+i \sin y$, $u=\cos z+i \sin z$, the equation $s+t+u=\frac{3 \sqrt{3}}{2}+i \frac{3}{2}$ in the complex numbers precisely means that the real part of the left side, $\cos x+\cos y+\cos z$, is equal to the real part of the right side, $\frac{3 \sqrt{3}}{2}$, and the imaginary part of the left side, $\sin x+\sin y+\sin z$, is equal to the imaginary part of the right side, $\frac{3}{2}$; these are precisely the equations of the problem.

Write the triangle inequality for the sum $s+t+u=\frac{3 \sqrt{3}}{2}+i \frac{3}{2}$:

$$
3=|s|+|t|+|u| \geq|s+t+u|=\left|\frac{3 \sqrt{3}}{2}+i \frac{3}{2}\right|=\sqrt{\left(\frac{3 \sqrt{3}}{2}\right)^{2}+\left(\frac{3}{2}\right)^{2}}=3
$$

The obtained equality precisely means that the vectors $s, t, u$ are parallel and equally oriented, so $|s|=|t|=|u|=1$ implies $s=t=u$, hence

$$
s=t=u=\frac{1}{3}\left(\frac{3 \sqrt{3}}{2}+i \frac{3}{2}\right)=\frac{\sqrt{3}}{2}+i \frac{1}{2}
$$

Therefore, $\cos x=\cos y=\cos z=\frac{\sqrt{3}}{2}, \sin x=\sin y=\sin z=\frac{1}{2}$, so the solution is:

$$
x=\frac{\pi}{6}+2 k \pi, \quad y=\frac{\pi}{6}+2 m \pi, \quad z=\frac{\pi}{6}+2 n \pi
$$

where $k, m, n$ are arbitrary integers.