Solve the following equation, where $n$ is a given number:

$$
\sin 3 x=\left(n^{2}-5 n+3\right) \sin x
$$

For which values of $n$ does this equation have (non-zero $\sin x$) solutions? Examine the cases when

$$
n^{2}-5 n+1=0
$$
$1^{0}$.

$$
\sin 3 x\left(n^{2}-5 n+3\right) \sin x \ldots
$$

the equation always has a solution $\sin x=0$. Indeed,

$$
\sin 3 x=3 \sin x-4 \sin ^{3} x, \quad \text { therefore } \quad\left(3-4 \sin ^{2} x\right) \sin x=\left(n^{2}-5 n+3\right) \sin x \ldots
$$

If $\sin x \neq 0$, we can simplify and thus obtain

$$
3-4 \sin ^{2} x=n^{2}-5 n+3 \quad \text { or } \quad \sin ^{2} x=\frac{n(5-n)}{4} \ldots
$$

The equation 3) has a valid solution if

$$
\begin{gathered}
00, \text { if } 0<n<5 \ldots \\
\frac{n(5-n)}{4} \leqq 1, \quad \text { if } \quad n(5-n) \leqq 4 \quad \text { or } \quad n^{2}-5 n+4 \leqq 0 \ldots
\end{gathered}
$$

Since $n^{2}-5 n+4=(n-1)(n-4)$, the condition 6) is satisfied if

$$
n<1 \quad \text { or } \quad n \geqq 4 \ldots
$$

Combining the conditions under 5) and 7) results in

$$
0<n \leqq 1 \quad \text { and } \quad 4 \leqq n<5 \ldots
$$

Accordingly, the equation 1) has solutions for values of $n$ that belong to the intervals given in 8). $2^{0}$. If $n^{2}-5 n+1=0$, or $n(5-n)=1$, then from 3)

$$
\sin ^{2} x=\frac{1}{4}, \quad \text { thus } \quad \sin x= \pm \frac{1}{2}
$$

Between 0 and $2 \pi$, the solutions are: $\frac{\pi}{6}, \pi-\frac{\pi}{6}, \pi+\frac{\pi}{6}$, and $2 \pi-\frac{\pi}{6}$.

In general: $2 k \pi+\frac{\pi}{6},(2 k+1) \pi-\frac{\pi}{6},(2 k+1) \pi+\frac{\pi}{6}, 2 \pi-\frac{\pi}{6}$.

These 4 sets of values can be written as $x=k \pi \pm \frac{\pi}{6}$, where $k$ is any integer.

Jozsef Klein (Izr. g. VIII. o. Debrecen)[^0]


[^0]:    ${ }^{1} \sin x=0$ in the case of $x=k \pi$.