Solve the equation $\sin 3 x=k \sin x$! What are the limits within which $k$ can vary so that this equation has solutions other than $\sin x=0$.

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I. Solution. Using the addition theorem:

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

If we substitute $\cos ^{2} x=1-\sin ^{2} x$ here, we get

$$
\sin 3 x=\left(3-4 \sin ^{2} x\right) \sin x=k \sin x
$$

an equation. Ignoring the solution $\sin x=0$,

$$
3-4 \sin ^{2} x=k, \quad \text { or } \quad \sin ^{2} x=\frac{3-k}{4}
$$

Since $\sin ^{2} x>0$, it must be that $k \leqq 3$; furthermore, it must be that

$\sin ^{2} x=\frac{3-k}{4} \leq 1$, i.e., $k \geq 1$

Thus, $k$ can be in the interval $[-1,+3]$.

If $k=-1, \sin ^{2} x=1$ and $\sin x= \pm 1$.

If $k=+3, \sin x=0$.

Horváth Sándor (Br. Kemény Zsigmond g. VI. o. Bp. VI.).

# II. Solution. 

$$
\sin 3 x=2 \sin x \cos ^{2} x+\cos 2 x \sin x=k \sin x
$$

from this equation, ignoring the solution $\sin x=0$, we get:

$$
2 \cos ^{2} x+\cos 2 x=k \quad \text { or } \quad 1+\cos 2 x+\cos 2 x=k
$$

and from this

$$
\cos 2 x=\frac{k-1}{2}
$$

Since $\cos 2 x \geq-1, \frac{k-1}{2} \geq-1$, if $k \geq-1$.

On the other hand, $\cos 2 x \leq+1$, so $\frac{k-1}{2} \leq+1$, if $k \leqq+3$.

Thus, there is a solution if $-1 \leq k \leq+3$.

Assuming that $k$ satisfies this condition, $\frac{k-1}{2}$ is the cosine of an angle that is between 0 and $\pi$; let such an angle be $\alpha$. Thus

$$
2 x= \pm \alpha+2 n \pi \text { or } x=n \pi \pm \frac{\alpha}{2}
$$

where $n$ represents any integer.