3. It is known that the numbers $x, y, z$ form an arithmetic progression in the given order with a common difference $\alpha=\arccos \frac{1}{9}$, and the numbers $5+\cos x, 5+\cos y, 5+\cos z$ form a non-constant geometric progression in the given order. Find $\cos y$.
Answer: $-\frac{1}{9}$.

Solution. Since the numbers $x, y, z$ form an arithmetic progression with a common difference $\alpha$, then $x=y-\alpha$, $z=y+\alpha$. Using the characteristic property of a geometric progression (the square of any term is equal to the product of its two neighbors), we obtain the equation

\[
\begin{gathered}
(5+\cos (y-\alpha))(5+\cos (y+\alpha))=(5+\cos y)^{2} \Leftrightarrow \\
\Leftrightarrow 5(\cos (y-\alpha)+\cos (y+\alpha))+\cos (y-\alpha) \cos (y+\alpha)=10 \cos y+\cos ^{2} y \Leftrightarrow \\
\Leftrightarrow 10 \cos y \cos \alpha+\frac{1}{2} \cos 2 \alpha+\frac{1}{2} \cos 2 y=10 \cos y+\cos ^{2} y \Leftrightarrow \\
\Leftrightarrow 10 \cos y(\cos \alpha-1)+\cos ^{2} \alpha+\cos ^{2} y-1=\cos ^{2} y \Leftrightarrow 10 \cos y(\cos \alpha-1)+(\cos \alpha-1)(\cos \alpha+1)=0
\end{gathered}
\]

From the last equation, it follows that either $\cos \alpha=1$, or $\cos y=-\frac{\cos \alpha+1}{10}$. The first case is impossible, since then $\alpha=2 \pi k, k \in \mathbb{Z}$, hence $\cos x=\cos y=\cos z$ and the geometric progression becomes constant. Therefore, the second case must hold. Since $\cos \alpha=\frac{1}{9}$, then $\cos y=-\frac{1}{9}$.