Solve the following system of equations:

$$
\begin{aligned}
\sin ^{2} x & =\sin y \\
\cos ^{4} x & =\cos y
\end{aligned}
$$
Let's square both equations and add the resulting two equations! Then on the right side, $\sin ^{2} y+\cos ^{2} y=1$, so we arrive at the equation

$$
\sin ^{4} x+\cos ^{8} x=1
$$

On the right side, $\sin ^{2} x+\cos ^{2} x$ can also be written, and thus we get the equation for $x$:

$$
\sin ^{4} x+\cos ^{8} x=\sin ^{2} x+\cos ^{2} x
$$

If $x \neq \frac{k \pi}{2}$ for some integer $k$, then $0<|\sin x|<1$ and $0<|\cos x|<1$, so $\sin ^{4} x<\sin ^{2} x$ and $\cos ^{8} x<\cos ^{2} x$, meaning the left side of (1) is definitely smaller than the right side. Therefore, (1) can only have solutions of the form $x=\frac{k \pi}{2}$, where $k$ is an integer. After this, we distinguish two cases:

I. Case. $k=2 l$, i.e., $x=l \pi$, where $l$ is an integer. Then $\sin ^{2} x=0, \cos ^{4} x=1$, so $\sin y=0, \cos y=1$, i.e., $y=2 m \pi$ for some integer $m$.

II. Case. $k=2 l+1$ odd, i.e., $x=l \pi+\frac{\pi}{2}$, where $l$ is an integer. Then $\sin ^{2} x=1, \cos ^{4} x=0$, so $\sin y=1, \cos y=0$, i.e., $y=2 m \pi+\frac{\pi}{2}$ for some integer $m$.

The solutions are therefore:

$$
\begin{array}{ll}
x=l \pi, & y=2 m \pi, \quad \text { where } l \text { and } m \text { are integers, or } \\
x=l \pi+\frac{\pi}{2}, & y=2 m \pi+\frac{\pi}{2}, \text { where } l \text { and } m \text { are integers. }
\end{array}
$$

It is easy to verify that these are indeed solutions to the system of equations.