2神 $\star$ Solve the system of equations: $\left\{\begin{array}{l}\cos x=2 \cos ^{3} y, \\ \sin x=2 \sin ^{3} y .\end{array}\right.$
Squaring and adding the two equations, we get
$$
\begin{aligned}
1 & =4\left(\cos ^{6} y+\sin ^{6} y\right)-4\left(\cos ^{4} y-\cos ^{2} y \sin ^{2} y+\sin ^{4} y\right) \\
& =4\left(1-3 \sin ^{2} y \cos ^{2} y\right)=4-3 \sin ^{2} 2 y .
\end{aligned}
$$

Thus, $\sin 2 y= \pm 1, y=\frac{k \pi}{2}+\frac{\pi}{4}(k \in \mathbf{Z})$. Substituting into the original system of equations, we get $x=2 l \pi+\frac{k \pi}{2}+\frac{\pi}{4}(l, k \in \mathbf{Z})$. Therefore, the solutions to the system of equations are
$$
\left\{\begin{array}{l}
x=2 l \pi+\frac{k \pi}{2}+\frac{\pi}{4}, \\
y=\frac{k \pi}{2}+\frac{\pi}{4} .
\end{array} \quad(l, k \in \mathbf{Z})\right.
$$