# Mathematical solution for intuitive trigonometric equation

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?

If triple angle relation for sine is known, letting $$\sin \alpha =s$$,
$$3 s -4 s^3= 2 s\rightarrow s =(0, \pm \frac12), \quad \alpha=(0, \pm 30^{\circ}, 150^{\circ}\pm30^{\circ} ) ..$$
$$\sin(3x) = \sin(2x + x) = \sin(2x)\cos(x) + \cos(2x)\sin(x).$$ 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?