Let $x$ represent a positive arc less than $2 \pi$ and $a$ be a given positive number. Determine the values of $x$ that satisfy the equation

$$
\sin 3 x + a \sin 2 x + 2 \sin x = 0
$$
The given equation can be written in the following form:

$$
(\sin 3 x+\sin x)+(a \sin 2 x+\sin x)=0
$$

The first term

$$
\begin{gathered}
\sin 3 x+\sin x=\sin 2 x \cos x+\cos 2 x \sin x+\sin x \\
=\sin 2 x \cos x+\sin x(1+\cos 2 x) \\
=\sin 2 x \cos x+\sin x\left(1+\cos ^{2} x-\sin ^{2} x\right) \\
=\sin 2 x \cos x+2 \sin x \cos ^{2} x \\
=2 \sin 2 x \cos x \\
=4 \sin x \cos ^{2} x
\end{gathered}
$$

The second term

$$
a \sin 2 x+\sin x=\sin x(1+2 a \cos x)
$$

and thus the entire equation takes the following form:

$$
\sin x\left(4 \cos ^{2} x+2 a \cos x+1\right)=0
$$

One solution is $\sin x=0$, from which

$$
x=0 \quad \text { or } \quad x=\pi
$$

On the other hand, the

$$
f(\cos x)=4 \cos ^{2} x+2 a \cos x+1=0
$$

equation provides two solutions for $x$ for each root that lies between -1 and +1, namely

$$
\alpha \quad \text { and } \quad 2 \pi-\alpha.
$$

which are positive and less than $2 \pi$.

For a root of the $f(\cos x)=0$ equation to satisfy this condition, it is necessary and sufficient that

$$
f(-1) f(+1) < 0
$$

which is equivalent to:

$$
a>\frac{5}{2}
$$

For both roots to satisfy the condition, the following relations must hold simultaneously:

$$
\begin{gathered}
(5-2 a)(5+2 a) \geqq 0 \\
4 a^{2}-16 \geqq 0 \\
-1<-\frac{a}{4}<1
\end{gathered}
$$

However, these conditions are only satisfied simultaneously if

$$
2 \leqq a \leqq \frac{5}{2}
$$

If $0<a<2$, the equation is only satisfied by the values $x=0$ and $x=\pi$.

*) See the article titled "Discussion of quadratic equations" in the first volume of "Középisk. Math. Lapok".