Solving a trigonometric equation I'm solving this equation:
$$\sin(3x) = 0$$
The angle is equal to 0, therefore:
$$3x=0+2k\pi \space\vee\space3x= (\pi-0)+2k\pi$$ 
$$x = \frac {2}{3}k\pi \space \vee \space x = \frac {\pi + 2k\pi}{3}$$
Though, the answer is 
$$x = k\frac {\pi}{3}$$
It looks like the two trigonometric equations have been combined into one. I must have made a mistake. Any hints?
 A: All we need is for $3x$ to be an integer multiple of $\pi$. In other words
$$3x = k\pi \Rightarrow x = \frac{k\pi}{3}$$
A: You're not missing anything: sometimes it's possible to efficiently combine sets of solutions.
Notice that for $\sin x=0$, the solutions $x=0+2k\pi$ ($0$ and then 'adding full circles') $\vee \; x=\pi+2k\pi$ ($\pi$ and then 'adding full circles') can be combined as $x=k\pi$ ($0$ and 'adding half circles'), where always $k \in \mathbb{Z}$.
Draw the solutions and realise that you're not 'missing' anything: both ways of writing down the solutions contain the exact same angles; you 'run through' the same angles.
Addendum
This is not always possible for equations of the form $\sin x = c$ (only if $c=k\pi$) or $\cos x = c$ (only if $c=\pi/2+k\pi$), but it is always possible for $\tan x = c$ since the solutions
$$x = \arctan c + 2k\pi \, \vee x = \pi + \arctan c + 2k\pi$$can alwayes be combined as
$$x = \arctan c + k\pi$$
You can easily see this by drawing a trigonometric circle and visualising the solutions.
A: Your solutions are:
$$
x = \frac{\pi}{3}(2k),\phantom{NNNNNNNN}
x = \frac{\pi}{3}(2k + 1).
$$
So, you've shown that $x$ is either $\frac{\pi}{3}$ times an even integer, or else $\frac{\pi}{3}$ times an odd integer.  In other words, $x$ is $\frac{\pi}{3}$ times an integer.
