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.
 A: Hint: The zeros of $\sin(x)$ occur at $k\pi$ for integers $k$. The zeros of $\cos(x)$ occur at $\frac{2k+1}{2}\pi$ for integers $k$. 

Full solution: The zeros of $\cos(x) = 0$ occur where $x = \frac{2k+1}{2}\pi$ for some integer $k$. And the zeros of $2\sin(2x) = 0$ occur where $2x = k\pi$ for some integer $k$, i.e. where $x = \frac{k\pi}{2}$ for some integer $k$. Thus the answer is (a). 

A: \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*}
A: How I would have approached this problem,
$$\sin 3x = 3\sin x - 4\sin^3x$$
$$\sin x + \sin 3x = 4\sin x - 4 \sin^3x = 0 \implies \sin x(1-\sin^2 x) = 0 \implies \sin x \cos^2x = 0$$
$$\implies x = n\pi \text{ or } x = n\pi+\pi/2$$
$$\implies x = 2n\frac{\pi}{2} \text{ or } \frac{(2n+1)\pi}{2} \implies x = k\frac{\pi}{2} \text{ or } \frac{(k+1)\pi}{2}$$
A: Hint:
$$
\sin \alpha =-\sin \beta \quad \iff \{\quad \alpha=-\beta+2n\pi \} \quad \mbox{or} \quad \{\alpha=\beta+(2n+1) \pi\}
$$
with$, \quad n \in \mathbb{Z}$
A: We need to solve 
$$\sin{x}=\sin(-3x),$$ 
which gives $x=-3x+2\pi k$ or $x=\pi-(-3x)+2\pi k$, where $k\in\mathbb Z$, which is
$x=\frac{\pi k}{2}$ or $x=-\frac{\pi}{2}+\pi k$, which gives the answer:
$$\left\{\frac{\pi k}{2}\big|k\in\mathbb Z\right\}$$
A: Factorizing $2\sin(2x)\cos(x)$ I got
$$4\cos(x)^2\sin(x)=0.$$ Therefore we get $\sin(x)=0$ or $\cos(x)=0$.
