Solve the following equation in the set of real numbers:

$$
\sin ^{2}(x+y)-\cos ^{2}(x-y)=1
$$
Solution. The functions $\sin ^{2} x$ and $\cos ^{2} x$ are always positive and at most 1, with their minimum value being 0. Therefore, the equation can only be satisfied if one of the functions has a value of 1 and the other has a value of 0.

1. Case. Let $\sin (x+y)=1$ and $\cos (x-y)=0$, then $x+y=\frac{\pi}{2}+2 k \pi, x-y=\frac{\pi}{2}+l \pi$, where $k$ and $l$ are integers. Adding the corresponding sides of the equations:

$$
x=\frac{\pi}{2}(2 k+l+1) \quad \text { and } \quad y=\frac{\pi}{2}(2 k-l) \text {. }
$$

2. Case. If $\sin (x+y)=-1$ and $\cos (x-y)=0$, then $x+y=-\frac{\pi}{2}+2 m \pi, x-y=\frac{\pi}{2}+n \pi$, where $m$ and $n$ are integers. The solution to the system of equations is:

$$
x=\frac{\pi}{2}(2 m+n), \quad y=\frac{\pi}{2}(2 m-n-1)
$$