Find those pairs of numbers $x, y$ for which

$$
\cos x + \cos y - \cos (x + y) = \frac{3}{2}
$$
I. solution. Using the known identities

$$
\cos x+\cos y=2 \cos \frac{x+y}{2} \cos \frac{x-y}{2} \quad \text { and } \quad \cos (x+y)=2 \cos ^{2} \frac{x+y}{2}-1
$$

our equation, after a simple transformation, can be written as:

$$
\left(2 \cos \frac{x+y}{2}-\cos \frac{x-y}{2}\right)^{2}+\sin ^{2} \frac{x-y}{2}=0
$$

which is equivalent to the following system of equations:

$$
\begin{gathered}
2 \cos \frac{x+y}{2}=\cos \frac{x-y}{2} \\
\sin \frac{x-y}{2}=0
\end{gathered}
$$

From (2), \( x-y=2 k \pi \) (where \( k \) is an integer), i.e., \( y=x-2 k \pi \). Substituting this into (1),

$$
2 \cos (x-k \pi)=\cos k \pi
$$

or

$$
2(-1)^{k} \cos x=(-1)^{k}
$$

thus

$$
\cos x=\frac{1}{2}
$$

and so the solutions to our equation are:

$$
x= \pm \frac{\pi}{3}+2 m \pi, \quad y= \pm \frac{\pi}{3}+2(m-k) \pi \quad(k, m \text { integers })
$$

where either both signs are plus or both are minus.

II. solution. Let \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \) be unit vectors originating from a common point such that the angles between \( \mathbf{a} \) and \( \mathbf{b} \), and between \( \mathbf{b} \) and \( \mathbf{c} \) are \( x+y \) and \( \pi-x \) respectively, and the angle between \( \mathbf{c} \) and \( \mathbf{a} \) is \( \pi-y \). Thus, based on the known properties of the dot product, for the vector \( \mathbf{v}=\mathbf{a}+\mathbf{b}+\mathbf{c} \),

$$
\begin{gathered}
\mathbf{v}^{2}=(\mathbf{a}+\mathbf{b}+\mathbf{c})^{2}=\mathbf{a}^{2}+\mathbf{b}^{2}+\mathbf{c}^{2}+2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a})= \\
3+2(\cos (x+y)+\cos (\pi-x)+\cos (\pi-y))= \\
3-2(\cos x+\cos y-\cos (x+y))
\end{gathered}
$$

thus

$$
\cos x+\cos y-\cos (x+y)=\frac{3-\mathbf{v}^{2}}{2} \leq \frac{3}{2}
$$

and equality holds if and only if \( \mathbf{v}=\mathbf{0} \), i.e., \( \mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0} \), which means that the three unit vectors form a regular triangle, so the absolute value of the angle between each pair is \( \frac{2 \pi}{3} \). Thus,

$$
\pi-x= \pm 2 \pi / 3+2 n \pi, \quad \pi-y= \pm 2 \pi / 3+2 m \pi
$$

( \( n, m \) integers), from which, after simple transformations, it can be seen that the solution is

$$
x= \pm \frac{\pi}{3}+2 k \pi, \quad y= \pm \frac{\pi}{3}+2 m \pi \quad(k, m \text { integers })
$$

Páles Zsolt (Sátoraljaújhely, Kossuth L. Gymnasium)