How many solutions does the equation

$$
\sin 2002 x=\sin 2003 x
$$

have in the interval $[0 ; 2 \pi]$?
Solution. The sine function is periodic with a period of $2 \pi$, so $x$ is certainly a solution if

$$
2002 x=2003 x+2 k \pi
$$

From this, $x=-2 k \pi$, but according to the condition $0 \leq x \leq 2 \pi$, so $k=0$ or -1. The solutions of the equation are $x=0$ and $x=2 \pi$. On the other hand, since $\sin \alpha=\sin (\pi-\alpha)$, the other solutions are obtained from the following equation:

$$
2002 x=\pi-2003 x+2 k \pi
$$

From this, $x=\frac{2 k+1}{4005} \pi$, and we still look for solutions in the interval $[0 ; 2 \pi]$, that is, from the inequality $0 \leq \frac{2 k+1}{4005} \leq 2$ we get $-0.5 \leq k \leq 4004.5$. Thus, the possible values of $k$ are $k=0,1,2, \ldots, 4004$.

Therefore, the equation has a total of 4007 different solutions, and these are:

$$
0, \quad \frac{1}{4005} \pi, \quad \frac{3}{4005} \pi, \quad \ldots, \quad \frac{2 k+1}{4005} \pi, \quad \ldots, \quad \frac{8009}{4005} \pi \quad \text { and } \quad 2 \pi
$$