11.3. Solve the equation: $|\sin x-\sin y|+\sin x \cdot \sin y=0$.
Answer: $x=\pi k, k \in Z ; y=\pi n, n \in Z$.

Solution. If $\sin x=0$, then $\sin y=0$, and vice versa, if $\sin y=0$, then $\sin x=0$. Therefore, $x=\pi k, k \in Z ; y=\pi n, n \in Z$. We will prove that there are no other solutions.

First method. Suppose $\sin x \neq 0$ and $\sin y \neq 0$. Then the given equation can have solutions if $\sin x \sin y < 0$. If $\sin x < 0, \sin y > 0$, we get the equation $(1-\sin x)(1+\sin y)=1$, which has no solution because the left side is always less than 1. Similarly, if $\sin x > 0, \sin y < 0$, we get the equation $(1+\sin x)(1-\sin y)=1$, which also has no solution for the same reason.

We can also transform the equation in different ways for each case where $\operatorname{sinxsin}y < 0, \operatorname{siny} < 0$ is considered similarly.

Second method. The equation is symmetric with respect to $\sin x$ and $\sin y$, so without loss of generality, we can assume that $\sin x \geq \sin y$. Then $\sin x - \sin y + \sin x \sin y = 0$. Further reasoning is similar to the above.

Evaluation criteria.

“+” A complete and well-reasoned solution is provided

“±” A generally correct reasoning is provided, containing minor gaps or inaccuracies

“P” Only the correct answer is provided

“-” The problem is not solved or solved incorrectly