Solve $\sin A +\sin 2A +\sin 3A + \sin 4A = 0$, for $0 \leq A \leq 180$ I've tried using factor formula but still did not manage to get the answer, not sure if factor formula is the right method.
I rearrange to $\sin 4A + \sin 2A + \sin 3A + \sin A = 0$,
and after applying factor formula,
$2 \sin 3A \cos A + 2 \sin 2A \cos A = 0$
$2 \cos A ( \sin 3A + \sin 2A) = 0$
$2 \cos A ( \sin \frac{5}{2} A \cos \frac{1}{2} A) = 0$
Then I'm stuck..

we have, $$\sin A+\sin 2A+\sin 3A+\sin 4A=0$$ $$(\sin A+\sin 3A)+(\sin 2A+\sin 4A)=0$$ $$2\sin\left(\frac{A+3A}{2}\right)\cos\left(\frac{A-3A}{2}\right)+2\sin\left(\frac{2A+4A}{2}\right)\cos \left(\frac{A-3A}{2}\right)=0$$ $$2\sin 2A\cos A+2\sin 3A\cos A=0$$ $$2\cos A(\sin 2A+\sin 3A)=0$$ $$\implies \cos A=0\implies \color{blue}{A=2n\pi+\frac{\pi}{2}}$$ $$\implies \sin 2A+\sin 3A=0\implies \sin 3A=-\sin 2A$$ $$\implies 3A=2n\pi-2A\implies \color{blue}{A=\frac{2n\pi}{5}}$$ Or $$3A=(2n+1)\pi-(-2A)$$ $$\implies \color{blue}{A=(2n+1)\pi}$$ Where, $n$ is any integer.
Edit: For $0\leq A\leq 180^\circ$ put $n=0$, $n=1$ & $n=2$ in the solutions, we get the following $$\color{blue}{A\in \left\{0, \frac{2\pi}{5}, \frac{\pi}{2},  \frac{4\pi}{5}, \pi\right\}}$$ or $$\color{blue}{A\in \left\{0^\circ, 72^\circ, 90^\circ, 144^\circ, 180^\circ\right\}}$$
