3. Determine all pairs of real numbers $x, y$ from the interval $[0,2 \pi]$ for which

$$
\sin x - \sin y + \cos (x - y) = \frac{3}{2}
$$
Solution.

Using the formulas $\sin \alpha-\sin \beta=2 \sin \frac{\alpha-\beta}{2} \cos \frac{\alpha+\beta}{2}$ and $\cos 2 \alpha=1-2 \sin ^{2} \alpha$, the equation can be transformed as follows:

$$
\begin{aligned}
& 2 \sin \frac{x-y}{2} \cos \frac{x+y}{2}+1-2 \sin ^{2} \frac{x-y}{2}=\frac{3}{2} \\
\Leftrightarrow & \sin ^{2} \frac{x-y}{2}-\sin \frac{x-y}{2} \cos \frac{x+y}{2}+\frac{1}{4}=0 \\
\Leftrightarrow & \left(\sin \frac{x-y}{2}-\frac{1}{2} \cos \frac{x+y}{2}\right)^{2}+\frac{1}{4}\left(1-\cos ^{2} \frac{x+y}{2}\right)=0
\end{aligned}
$$

Both addends are non-negative, and their sum is 0, so it must be

$$
\begin{aligned}
& 1-\cos ^{2} \frac{x+y}{2}=0 \\
& \sin \frac{x-y}{2}-\frac{1}{2} \cos \frac{x+y}{2}=0
\end{aligned}
$$

We distinguish two cases:

(i) $\cos \frac{x+y}{2}=1, \sin \frac{x-y}{2}=\frac{1}{2}$

Because $0 \leq \frac{x+y}{2} \leq 2 \pi$ and $\cos \frac{x+y}{2}=1$, it must be $x=y=0$ or $x=y=2 \pi$, which is in contradiction with the second equation.

(ii) $\cos \frac{x+y}{2}=-1, \sin \frac{x-y}{2}=-\frac{1}{2}$

Because $0 \leq \frac{x+y}{2} \leq 2 \pi$ and $\cos \frac{x+y}{2}=-1$, it must be $\frac{x+y}{2}=\pi$, i.e., $y=2 \pi-x$. Substituting into the second equation gives $\sin (x-\pi)=-\frac{1}{2}$, i.e., $\sin x=\frac{1}{2}$. We obtain the solutions

$$
x=\frac{\pi}{6}, y=\frac{11 \pi}{6} \quad \text { and } \quad x=\frac{11 \pi}{6}, y=\frac{\pi}{6} .
$$