Mathematical solution for $\sin 3\alpha = 2\sin\alpha$ where $\alpha$ is acute

Question

I derived the following trigonometrical equation from a real triangle, knowing that the angle $\alpha$ is an acute one:

$\sin 3\alpha = 2\sin\alpha$

Just by eye-balling the equation, and remembering the trigonometric unit circle, I know that:

$\sin 90 = 1 = 2\sin 30 = 2 \cdot 0.5 = 1$

Therefore, $\alpha = 30^\circ$ is a possible solution.
I am unaware of any trigonometric identity to help me simplify this equation in order to get all possible solutions for $\alpha$, and this intuitive solution is the best I can come up with. I plugged this equation into symbolab.com, but their solution seems very long-winded, and I am hoping for the possibility that a simpler one exists.

How can I solve this type of problem when the intuitive approach fails?

Answer 1

$$\sin 3\alpha = 2\sin\alpha$$

\begin{align} \sin(3\alpha) &= \sin(2\alpha + \alpha) \\ &= \sin(2\alpha) \cos \alpha + \cos(2 \alpha) \sin \alpha \\ &= (2 \sin \alpha \cos \alpha) \cos \alpha + (\cos^2 \alpha -\sin^2 \alpha) \sin \alpha \\ &= 2\sin \alpha \cos^2\alpha + \cos^2\alpha \sin \alpha - \sin^3 \alpha \\ &= 3\sin \alpha \cos^2 \alpha -\sin^3 \alpha \\ &= 3 \sin \alpha(1 - \sin^2 \alpha) - \sin^3 \alpha \\ &= 3\sin \alpha - 4 \sin^3 \alpha \end{align}

\begin{align} \sin 3\alpha &= 2\sin\alpha \\ 3\sin \alpha - 4 \sin^3 \alpha &= 2\sin \alpha \\ \sin \alpha - 4\sin^3 \alpha &= 0 \\ \sin \alpha (1 - 4 \sin^2 \alpha) &= 0 \\ \hline \sin \alpha &= 0 \\ \sin^2 \alpha &= \dfrac 14 \\ \hline \sin \alpha &\in \left\{-\dfrac 12, 0, \dfrac 12 \right\} \end{align}

Answer 2

If triple angle relation for sine is known, letting $ \sin \alpha =s$,

$$3 s -4 s^3= 2 s\rightarrow s =(0, \pm \frac12), \quad \alpha=(0, \pm 30^{\circ}, 150^{\circ}\pm30^{\circ} ) ..$$

Answer 3

Here is a lead.

$$\sin(3x) = \sin(2x + x) = \sin(2x)\cos(x) + \cos(2x)\sin(x).$$ Now invoke some double angle magig to get $$\sin(3x) = 2\sin(x)\cos^2(x) + (1 - 2\sin^2(x))\sin(x).$$ Next, use the pythagorean identities to gete $$\sin(3x) = 2\sin(x)(1 - \sin^2(x)) + (1 - 2\sin^2(x))\sin(x). $$ Can you use this?