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..
 A: From the point that you stopped

$$2 \cos A ( \sin \frac{5}{2} A \cos \frac{1}{2} A) = 0$$

$$2\cos(A/2)\cos(A)\sin(\frac{5A}{2})=0 \bigg/:2$$
$$\cos(A/2)\cos(A)\sin(\frac{5A}{2})=0$$
$\cos(A/2)=0\;$ or $\cos(A)=0\;$ or $\sin(\frac{5A}{2})=0$
A: You've done the hard work
Now the product of three multiplicands is zero
so at least one of them must be equal to zero
If $\sin B=0,B=n180^\circ$
If $\cos C=0, C=(2m+1)90^\circ$ where $m,n$ are integers
A: Had the number of summands been larger, we could employ the method dsecribed in How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?
If $\sin\dfrac A2=0,$ given expression holds true 
and $\dfrac A2=n180^\circ\iff A=n360^\circ$ where $n$ is any integer
Else $A\ne n360^\circ$ and $\sin A+\sin2A+\sin3A+\sin4A=\cdots=\dfrac{\cos\dfrac A2-\cos\dfrac{9A}2}{\sin\dfrac A2}$
So we need $\cos\dfrac A2-\cos\dfrac{9A}2=0\iff\cos\dfrac{9A}2=\cos\dfrac A2$
$\implies\dfrac{9A}2=360^\circ m\pm\dfrac A2$ where $m$ is any integer, but $A\ne n360^\circ$
Consider the +, - separately 
A: 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 $$\color{blue}{A\in \left\{0^\circ, 72^\circ, 90^\circ, 144^\circ, 180^\circ\right\}}$$
A: I solve and got 0, 90, 144, 180 but im still missing 72(answer sheet shows that 72 is one of the answer). 
Can someone help me get 72 ?
A: To be honest I actually don't understand what are y'all doing, maybe a little but mostly no.
here's how I approach,
Let basic angle = a
cos A = 0
  a = 90
A = 90 
sin 5/2 A = 0 
a = 0
5/2 A = 0, 360
A = 0, 144 
cos 1/2 A = 0
a = 90
1/2 A = 90
A = 180
Therefore, A = 0, 90, 144, 180
But where's the 72 T.T
