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.

\begin{align*}\sin (2x)=0 &\Longrightarrow 2x=k\pi\to x=\frac{k\pi}{2}\\
\cos x=0&\Longrightarrow x= \frac{\pi}{2}+k\pi=\frac{(2k+1)\pi}{2}\end{align*}
