2. It is known that the numbers $x, y, z$ form an arithmetic progression in the given order with a common difference $\alpha=\arccos \frac{2}{3}$, and the numbers $\frac{1}{\sin x}, \frac{6}{\sin y}, \frac{1}{\sin z}$ also form an arithmetic progression in the given order. Find $\sin ^{2} y$.
Answer: $\frac{5}{8}$.

Solution. Using the characteristic property of an arithmetic progression, we get $y=\frac{x+z}{2}$, $\frac{12}{\sin y}=\frac{1}{\sin x}+\frac{1}{\sin z}$. Transform the second equation:

$$
\frac{12}{\sin y}=\frac{\sin x+\sin z}{\sin x \sin z} \Leftrightarrow \frac{12}{\sin y}=\frac{2 \sin \frac{x+z}{2} \cos \frac{x-z}{2}}{0.5(\cos (x-z)-\cos (x+z))}
$$

Substituting here that $x+z=2 y, x-z=-2 \alpha$, we get $\frac{3}{\sin y}=\frac{\sin y \cos \alpha}{\cos 2 \alpha-\cos 2 y}$. This equality is equivalent to the following on the domain of definition:

$$
3-6 \sin ^{2} \alpha-3+6 \sin ^{2} y=\sin ^{2} y \cos \alpha \Leftrightarrow(6-\cos \alpha) \sin ^{2} y=6 \sin ^{2} \alpha \Leftrightarrow \sin ^{2} y=\frac{6 \sin ^{2} \alpha}{6-\cos \alpha}
$$

Since $\alpha=\arccos \frac{2}{3}$, then $\sin ^{2} \alpha=\frac{5}{9}, \cos \alpha=\frac{2}{3}$, and finally we get $\sin ^{2} y=\frac{5}{8}$.