Characterization of the $x$ such that $\sin(x)$ is rational? For $x \in [0,\pi/2]$, $\sin(x)$ ranges over $[0,1]$. 
So every rational number in $[0,1]$ is the sine of some $x \in [0,\pi/2]$.

Q. Is there any characterization of the $x$ for which $\sin(x)$ is rational?

I am not quite sure what shape such a characterization might take.
Something like: If $x$ satisfies such-and-such conditions, then $\sin(x)$ is rational. Perhaps there is at least a partial characterization?
 A: First note that:
$x\to \sin(x)$ is locally a bijection on the interval you have given , so it's rational for an infinite number of arguments, in particular whenever it's evaluated on $\arcsin(r)$ for rational $r$
Another way of solving your problem is assuming that $\sin(x)$ and $\cos(x)$ are together rationals and here we will come up with the Pythagorean triples to solve the problem,if $$
(x,y)=\left( \frac{n^2-m^2}{n^2+m^2}, \frac{2mn}{n^2+m^2} \right).
$$
then $x^2+y^2=1$, so it's on the unit circle meaning that it's a value for some$(\sin(t),\cos(t))$
Finally, According to Niven's theorem in analysis which says that $\sin(r\pi)=q$ is rational if and only if $q=0,1,-1,\mp\frac{1}{2}$
A: The only nice $x$ in this range for which $\sin x$ is rational are $0$, $\pi/6$, and $\pi/2$. Every other preimage of a rational number is neither rational nor a rational multiple of $\pi$.
(In fact I'm almost sure that if $\sin x$ is rational and not in $\{-1,-\frac12,0,\frac12,1\}$, then $x$ is algebraically independent of $\pi$).
A: I don't know if this is what you are looking for, but if we work on the unit circle, $sinx$ is the y-coordinate of any point $(x,y)$. Then $a^2+b^2=1$ , where $a=cosx, b=sinx$ . Then $$ b=\sqrt{1-a^2} \in \mathbb Q $$ iff $(1-a^2)= \frac {p^2}{q^2}$ , so $y$ as the coordinate of a point is rational iff the x coordinate is of the form $ \sqrt { 1-p^2/q^2}$ 
Or take any Pythagorean triple $(a,b,c)$  so that $a^2+b^2=c^2$ . Dividing through by $c$, you get $(a/c)^2+(b/c)^2 =1$ Every such triple (of course, (a,b,c) relatively prime) will give you a Rational sine (and a Rational cosine). These are actually all your Rational sines.
A: $\sin(x)=$ opposite/hypotenuse.
$y^2+x^2=1$ 
If $\frac{y}{y^2+x^2}$ is rational.
So we know $y$ must be rational and consider: $y^2+x^2$? what conditions would make $y^2+x^2$ rational?
(fixed errors)
