## Task A-4.1. (10 points)

Real numbers $x, y$ and $z$ are consecutive terms of an arithmetic sequence. If the numbers $\cos ^{2} x$, $\cos ^{2} y$ and $\cos ^{2} z$ are distinct and consecutive terms of some arithmetic sequence, determine all possible values of $y$.
## First Solution.

Since $x, y$ and $z$ are consecutive terms of an arithmetic sequence, there exists $t \in \mathbb{R}$ such that $x=y-t$ and $z=y+t$.

The condition that $\cos ^{2} x, \cos ^{2} y$ and $\cos ^{2} z$ are consecutive terms of an arithmetic sequence can be written as

$$
\cos ^{2} x+\cos ^{2} z=2 \cos ^{2} y
$$

Now we have

$$
\begin{gathered}
\cos ^{2}(y-t)+\cos ^{2}(y+t)=2 \cos ^{2} y \\
(\cos y \cos t+\sin y \sin t)^{2}+(\cos y \cos t-\sin y \sin t)^{2}=2 \cos ^{2} y \\
\cos ^{2} y \cos ^{2} t+2 \sin y \cos y \sin t \cos t+\sin ^{2} y \sin ^{2} t+ \\
\cos ^{2} y \cos ^{2} t-2 \sin y \cos y \sin t \cos t+\sin ^{2} y \sin ^{2} t=2 \cos ^{2} y \\
\cos ^{2} y \cos ^{2} t+\sin ^{2} y \sin ^{2} t=\cos ^{2} y
\end{gathered}
$$

from which it follows that

$$
\begin{gathered}
\sin ^{2} y \sin ^{2} t=\cos ^{2} y\left(1-\cos ^{2} t\right) \\
\sin ^{2} y \sin ^{2} t=\cos ^{2} y \sin ^{2} t \\
\sin ^{2} t\left(\cos ^{2} y-\sin ^{2} y\right)=0
\end{gathered}
$$

This is possible only if $\sin t=0$ or $\cos y= \pm \sin y$.

If $\sin t=0$, then $t=k \pi$, for $k \in \mathbb{Z}$. In this case, $x=y-k \pi, z=y+k \pi$ and thus $\cos ^{2} x=\cos ^{2} z=\cos ^{2} y$. Therefore, this case is invalid.

2 points

Thus, $\cos y= \pm \sin y$, or $\operatorname{tg} y= \pm 1$. Finally, $y \in\left\{\left.\frac{\pi}{4}+k \cdot \frac{\pi}{2} \right\rvert\, k \in \mathbb{Z}\right\}$.