2. It is known that $\frac{\sin 3 x}{(2 \cos 2 x+1) \sin 2 y}=\frac{1}{5}+\cos ^{2}(x-2 y)$ and $\frac{\cos 3 x}{(1-2 \cos 2 x) \cos 2 y}=\frac{4}{5}+\sin ^{2}(x-2 y)$. Find all possible values of the expression $\cos (x-6 y)$, given that there are at least two. Answer: 1 or $-\frac{3}{5}$.
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

$$
\left\{\begin{array}{l}
\frac{\sin x}{\sin 2 y}=\frac{1}{5}+\cos ^{2}(x-2 y) \\
-\frac{\cos x}{\cos 2 y}=\frac{4}{5}+\sin ^{2}(x-2 y)
\end{array}\right.
$$

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

$$
\sin (x-2 y)=\sin 4 y \Leftrightarrow\left[\begin{array} { l } 
{ x - 2 y = 4 y + 2 \pi k } \\
{ x - 2 y = \pi - 4 y + 2 \pi k , k \in \mathbb { Z } }
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=6 y+2 \pi k \\
x=\pi-2 y+2 \pi k, k \in \mathbb{Z}
\end{array}\right.\right.
$$

Let's consider these two possibilities separately.

If $x=\pi-2 y+2 \pi k, k \in \mathbb{Z}$, then the system (??) takes the form $\left\{\begin{array}{l}1=\frac{1}{5}+\cos ^{2} 4 y, \\ 1=\frac{4}{5}+\sin ^{2} 4 y ;\end{array} \quad\right.$ from which $\cos ^{2} 4 y=\frac{4}{5}, \sin ^{2} 4 y=\frac{1}{5}$. It is easy to see that this system has solutions. In this case, $\cos (x-6 y)=$ $-\cos 8 y=-2 \cos ^{2} 4 y+1=-2 \cdot \frac{4}{5}+1=-\frac{3}{5}$.

If $x=6 y+2 \pi k, k \in \mathbb{Z}$, then (??) takes the form

$$
\left\{\begin{array} { l } 
{ \frac { \sin 6 y } { \sin 2 y } = \frac { 1 } { 5 } + \cos ^ { 2 } 4 y , } \\
{ - \frac { \cos 6 y } { \cos 2 y } = \frac { 4 } { 5 } + \sin ^ { 2 } 4 y }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
3-4 \sin ^{2} 2 y=\frac{1}{5}+\cos ^{2} 4 y \\
3-4 \cos ^{2} 2 y=\frac{4}{5}+\sin ^{2} 4 y
\end{array}\right.\right.
$$

Each of the equations in the last system reduces to the equation $\cos ^{2} 4 y-2 \cos 4 y-\frac{4}{5}=0$, which has solutions, since $\cos 4 y=1-\frac{3}{\sqrt{5}}$. Therefore, this case is also possible, and we get that $\cos (x+6 y)=\cos (2 \pi k)=1$.

Thus, $\cos (x-6 y)$ can be equal to $-\frac{3}{5}$ or 1.

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