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

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

\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}