Trig equation $\sin(x)+\sin(3x)=0$, the answers are given in a factored form $\sin(x)+\sin(3x)=0$
So to solve this I tried the following:
-First I transformed this expression from sum to product because it equals zero
$\sin(x)+\sin(3x)=0 => 2\sin(\frac{x+3x}{2})\cdot\cos(\frac{x-3x}{2})=0$ 
-Second part I got to 
$2\sin(2x)\cos(x)=0$
-Third part:
I have separated the equation into two parts:
$2\sin(2x)=0$   and   $\cos(x)=0$ 
From now on I don't know how to get to the correct answers, which option should i circle and why:
a) $x=(2k+1)\frac{\pi}{2}$ or $x=\frac{k\pi}{2}$
b) $x=(k+1)\frac{\pi}{2}$ or $x=\frac{k\pi}{2}$
c) $x=(3k+1)\frac{\pi}{3}$ or $x=\frac{3k\pi}{2}$
d) $x=(k-1)\frac{\pi}{2}$ or $x=\frac{3k\pi}{2}$
where $k$ is any integer.

How I would have approached this problem,
$$\sin 3x = 3\sin x - 4\sin^3x$$
$$\sin x + \sin 3x = 4\sin x - 4 \sin^3x = 0 \implies \sin x(1-\sin^2 x) = 0 \implies \sin x \cos^2x = 0$$
$$\implies x = n\pi \text{ or } x = n\pi+\pi/2$$
$$\implies x = 2n\frac{\pi}{2} \text{ or } \frac{(2n+1)\pi}{2} \implies x = k\frac{\pi}{2} \text{ or } \frac{(k+1)\pi}{2}$$
