Solve the following equation, where $p$ and $q$ are odd integers:

$\sin p x \cdot \sin q x+2 \sin p x-3 \sin q x-4=0$.
Subtracting two from both sides of the equation, the left side can be factored as:

$$
(\sin p x-3)(\sin q x+2)=-2
$$

Since $-4 \leqq \sin p x-3 \leqq-2$ and $1 \leqq \sin q x+2 \leqq 3$, the absolute value of the left side can only be 2 if

$$
\sin p x-3=-2 \quad \text { and } \quad \sin q x+2=1
$$

which means

$$
p x=\frac{\pi}{2}(4 n+1) \quad \text { and } \quad q x=\frac{\pi}{2}(4 m+3)
$$

for suitable integers $n$ and $m$. We still need to examine for which integers $n$ and $m$ the system of equations (1) has a solution. Clearly, we get systems of equations equivalent to (1) if we replace the second equation with the ratio of the two original equations:

$$
x=\frac{\pi}{2 p}(4 n+1), \quad \frac{p}{q}=\frac{4 n+1}{4 m+3}
$$

However, this is solvable if and only if

$$
\frac{p}{q}=\frac{4 n+1}{4 m+3}
$$

Let the greatest common divisor of $p$ and $q$ be $d$, then $p=d p_{1}$ and $q=d q_{1}$, so

$$
\frac{p_{1}}{q_{1}}=\frac{4 n+1}{4 m+3}
$$

The fraction $\frac{p_{1}}{q_{1}}$ is already in its simplest form, so ( $3^{\prime}$ ') is only satisfied if for some integer $k$,

$$
4 n+1=k p_{1} \quad \text { and } \quad 4 m+3=k q_{1} \text {. }
$$

For solvability, it is necessary that $p_{1}$ and $q_{1}$ give different remainders when divided by 4, hence - since both are odd - one gives a remainder of 1 and the other -1. The integer $k$ is also odd, so ( $4 n+1=k p_{1}$ implies) it gives the same remainder when divided by 4 as $p_{1}$, that is,

$$
k=4 t+p_{1}
$$

Conversely, if we choose the value of $k$ according to (5) (for any integer $t$), then

$$
\begin{aligned}
k p_{1} & =4 t p_{1}+p_{1}^{2}=4 t p_{1}+(4 v \pm 1)^{2}= \\
& =4\left(t p_{1}+4 v^{2} \pm 2 v\right)+1, \text { or } \\
k q_{1} & =\left(p_{1}+q_{1}\right) k-k p_{1}=4 s k-k p_{1}= \\
& =4\left(s k-t p_{1}-4 v^{2} \pm 2 v\right)-1
\end{aligned}
$$

indeed gives the desired remainder when divided by 4; thus (4) is solvable, and so is the original system of equations (1). Based on (2), (4), and (5), the solution is:

$$
x=\frac{\pi}{2} \cdot \frac{4 n+1}{p}=\frac{\pi}{2} \cdot \frac{k}{d}=\frac{\pi}{2} \cdot \frac{4 t+p_{1}}{d}
$$

Remarks. 1. Many of the submitters did not notice that the two equations for $x$ do not always have a common solution (e.g., when $p=q=1$), and, if a solution does exist, the value of $n$ cannot be any integer.

2. Many reached the point where $\sin p x=1$ and $\sin q x=-1$, from which they concluded the vanishing of one of the factors from the equation $0=\sin p x+\sin q x=2 \sin \frac{p+q}{2} x \cos \frac{p+q}{2} x$. However, this way they also obtained false roots, since $\sin p x+\sin q x$ can be zero not only when $\sin p x=1$ and $\sin q x=-1$.