Problem 949

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)