4. 119 Solve the system of equations $\left\{\begin{array}{l}\cos x=2 \cos ^{3} y, \\ \sin x=2 \sin ^{3} y .\end{array}\right.$
[Solution] Squaring and adding the two equations, we get
$$\begin{aligned}
1 & =4\left(\cos ^{6} y+\sin ^{6} y\right) \\
& =4\left(\cos ^{4} y-\cos ^{2} y \sin ^{2} y+\sin ^{4} y\right) \\
& =4\left(1-3 \sin ^{2} y \cos ^{2} y\right)
\end{aligned}$$

So $\sin 2 y= \pm 1, y=\frac{k \pi}{2}+\frac{\pi}{4}(k \in \mathbb{Z})$, substituting into the original system of equations gives $\quad x=2 l \pi+\frac{k \pi}{2}+\frac{\pi}{4}(l, k \in \mathbb{Z})$.

Therefore, the solution to the system of equations is
$$\left\{\begin{array}{l}
x=2 l \pi+\frac{k \pi}{2}+\frac{\pi}{4}, \\
y=\frac{k \pi}{2}+\frac{\pi}{4} .
\end{array} \quad(l, k \in \mathbb{Z})\right.$$