Problem 11.3. Let $a$ be an integer. Find all real numbers $x, y, z$ such that

$$
a(\cos 2 x+\cos 2 y+\cos 2 z)+2(1-a)(\cos x+\cos y+\cos z)+6=9 a
$$

Aleksandar Ivanov
11.3. Using the formula $\cos 2 \alpha=2 \cos ^{2} \alpha-1$, the equation becomes

$$
a\left(\cos ^{2} x+\cos ^{2} y+\cos ^{2} z\right)+(1-a)(\cos x+\cos y+\cos z)+3-6 a=0
$$

Consider the function $f(t)=a t^{2}+(1-a) t+1-2 a, t \in[-1,1]$. The roots of the equation $f(t)=0$ are $t_{1}=-1$ and $t_{2}=\frac{2 a-1}{a}, a \neq 0$. The following three cases are possible:

Case 1. $a1$, it follows that $f(t) \geq 0$ for any $t \in[-1,1]$ and $f(t)=0$ if and only if $t=-1$.

Case 2. $a=0$. Then $f(t)=t+1 \geq 0$ for any $t \in[-1,1]$ and $f(t)=0$ if and only if $t=-1$.

Case 3. $a>0$. Then $a \geq 1$ and hence $\frac{2 a-1}{a} \geq 1$ with equality for $a=1$. It follows that $f(t) \geq 0$ for any $t \in[-1,1]$. Moreover, if $a>1$ then $f(t)=0$ for $t=-1$, and if $a=1$, then $f(t)=0$ for $t= \pm 1$.

Since the given equation has the form $f(\cos x)+f(\cos y)+f(\cos z)=0$, we conclude that:

- if $a \neq 1$, then $\cos x=\cos y=\cos z=-1$. Hence the solutions of the problem are $x=(2 k+1) \pi, y=(2 l+1) \pi, z=(2 m+1) \pi$, where $k, l, m \in \mathbb{Z}$.
- if $a=1$, then in addition to the above solutions we also have $\cos x=$ $\cos y=\cos z=1$, that is, $x=2 r \pi, y=2 s \pi, z=2 t \pi$, where $r, s, t \in \mathbb{Z}$.