2. It is known that $\frac{\cos 3 x}{(2 \cos 2 x-1) \cos y}=\frac{2}{5}+\cos ^{2}(x+y)$ and $\frac{\sin 3 x}{(2 \cos 2 x+1) \sin y}=\frac{3}{5}+\sin ^{2}(x+y)$. Find all possible values of the expression $\cos (x+3 y)$, given that there are at least two.
Answer: -1 or $\frac{1}{5}$.

Solution. Note that

$$
\begin{aligned}
& \cos 3 x=\cos (2 x+x)=\cos 2 x \cos x-\sin 2 x \sin x=\cos 2 x \cos x-2 \sin ^{2} x \cos x= \\
& \quad=\cos 2 x \cos x-(1-\cos 2 x) \cos x=(2 \cos 2 x-1) \cos x \\
& \sin 3 x=\sin (2 x+x)=\sin 2 x \cos x+\sin x \cos 2 x= \\
& \quad=2 \sin x \cos ^{2} x+\sin x \cos 2 x=\sin x(1+\cos 2 x)+\sin x \cos 2 x=(2 \cos 2 x+1) \sin x
\end{aligned}
$$

Taking this into account, the given equalities take the form

$$
\left\{\begin{array}{l}
\frac{\cos x}{\cos y}=\frac{2}{5}+\cos ^{2}(x+y) \\
\frac{\sin x}{\sin y}=\frac{3}{5}+\sin ^{2}(x+y)
\end{array}\right.
$$

(it is also necessary to consider that $\cos 2 x \neq \pm \frac{1}{2}$). Adding the equations (??) we get $\frac{\cos x}{\cos y}+$ $\frac{\sin x}{\sin y}=2$, from which $\cos y \neq 0, \sin y \neq 0$ and $\cos x \sin y+\sin x \cos y=2 \sin y \cos y$. The last equality is equivalent to the following:

$$
\sin (x+y)=\sin 2 y \Leftrightarrow\left[\begin{array}{l}
x + y = 2 y + 2 \pi k \\
x + y = \pi - 2 y + 2 \pi k , k \in \mathbb { Z }
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=y+2 \pi k \\
x=\pi-3 y+2 \pi k, k \in \mathbb{Z}
\end{array}\right.\right.
$$

Let's consider these two possibilities separately

If $x=y+2 \pi k, k \in \mathbb{Z}$, then the system (??) takes the form $\left\{\begin{array}{l}1=\frac{2}{5}+\cos ^{2} 2 y, \\ 1=\frac{3}{5}+\sin ^{2} 2 y ;\end{array} \quad\right.$ from which $\sin ^{2} 2 y=\frac{2}{5}$, $\cos ^{2} 2 y=\frac{3}{5}$. It is easy to see that this system has solutions. In this case, $\cos (x+3 y)=\cos 4 y=$ $2 \cos ^{2} 2 y-1=2 \cdot \frac{3}{5}-1=-\frac{1}{5}$.

If $x=\pi-3 y+2 \pi k, k \in \mathbb{Z}$, then (??) takes the form

$$
\left\{\begin{array}{l}
- \frac { \cos 3 y } { \cos y } = \frac { 2 } { 5 } + \cos ^ { 2 } 2 y , \\
\frac { \sin 3 y } { \sin y } = \frac { 3 } { 5 } + \sin ^ { 2 } 2 y
\end{array} \Leftrightarrow \left\{\begin{array}{l}
3-4 \cos ^{2} y=\frac{2}{5}+\cos ^{2} 2 y \\
3-4 \sin ^{2} y=\frac{3}{5}+\sin ^{2} 2 y
\end{array}\right.\right.
$$

Each of the equations in the last system reduces to the equation $\cos ^{2} 2 y+2 \cos 2 y-\frac{3}{5}=0$, which has solutions, since $\cos 2 y=-1+\sqrt{\frac{8}{5}}$. Therefore, this case is also possible, and we get that $\cos (x+3 y)=\cos (\pi+2 \pi k)=-1$.

Thus, $\cos (x+3 y)$ can be equal to $\frac{1}{5}$ or -1.

Note that in both cases all the restrictions that appeared in connection with the domain of definition $\left(\cos 2 x \neq \pm \frac{1}{2}\right.$, $\cos y \neq 0, \sin y \neq 0$ ) are satisfied.

Remark. Since the problem states that the expression $\cos (x+6 y)$ has at least two different values, checking the possibility of the cases is not mandatory.