3. It is known that the numbers $x, y, z$ form an arithmetic progression in the given order with a common difference $\alpha=\arccos \left(-\frac{1}{5}\right)$, and the numbers $2+\sin x, 2+\sin y, 2+\sin z$ form a non-constant geometric progression in the given order. Find $\sin y$.
Answer: $-\frac{1}{5}$.

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 the two adjacent terms), we obtain the equation

$$
\begin{gathered}
(2+\sin (y-\alpha))(2+\sin (y+\alpha))=(2+\sin y)^{2} \Leftrightarrow \\
\Leftrightarrow 2(\sin (y-\alpha)+\sin (y+\alpha))+\sin (y-\alpha) \sin (y+\alpha)=4 \sin y+\sin ^{2} y \Leftrightarrow \\
\Leftrightarrow 4 \sin y \cos \alpha+\frac{1}{2} \cos 2 \alpha-\frac{1}{2} \cos 2 y=4 \sin y+\sin ^{2} y \Leftrightarrow \\
\Leftrightarrow 4 \sin y(\cos \alpha-1)+\cos ^{2} \alpha-\cos ^{2} y=\sin ^{2} y \Leftrightarrow 4 \sin y(\cos \alpha-1)+(\cos \alpha-1)(\cos \alpha+1)=0
\end{gathered}
$$

From the last equation, it follows that either $\cos \alpha=1$, or $\sin y=-\frac{\cos \alpha+1}{4}$. The first case is impossible, since then $\alpha=2 \pi k, k \in \mathbb{Z}$, hence $\sin x=\sin y=\sin z$ and the geometric progression turns out to be constant. Therefore, the second case must hold. Since $\cos \alpha=-\frac{1}{5}$, then $\sin y=-\frac{1}{5}$.