2. It is known that the numbers $x, y, z$ form an arithmetic progression in the given order with a common difference $\alpha=\arccos \left(-\frac{1}{3}\right)$, and the numbers $\frac{1}{\cos x}, \frac{3}{\cos y}, \frac{1}{\cos z}$ also form an arithmetic progression in the given order. Find $\cos ^{2} y$.
Answer: $\frac{4}{5}$.

Solution. Using the characteristic property of an arithmetic progression, we get $y=\frac{x+z}{2}$, $\frac{14}{\cos y}=\frac{1}{\cos x}+\frac{1}{\cos z}$. Transform the second equation:

$$
\frac{6}{\cos y}=\frac{\cos x+\cos z}{\cos x \cos z} \Leftrightarrow \frac{6}{\cos y}=\frac{2 \cos \frac{x+z}{2} \cos \frac{x-z}{2}}{0.5(\cos (x+z)+\cos (x-z))}
$$

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

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

Since $\cos \alpha=-\frac{1}{3}$, we finally get $\cos ^{2} y=\frac{4}{5}$.