I'm a little confused about how to find Laplace transforms of a sine function when it is a function of time.
As in, suppose the function is $x(t)=\sin(at)$ , then I can proceed to get $X(s)=a/(s^2+a^2)$.
However, now I have a function $\overset{.}x(t) = \sin(x(t))$ [LHS is derivative of $x(t)$ wrt. $t$]. How do I find the Laplace transform of this?
Taking the Laplace transform of both members gives you:
$sX(s)-X_0=\int_0^\infty\sin(x(t))\ e^{-st}dt$
and that's about all you can say as there is no rule for a nonlinear transform of the signal.
You could expand the sine function using McLaurin, and use the formula for function multiplication to evaluate the transforms of the powers of $x(t)$ (http://en.wikipedia.org/wiki/Laplace_transform#Properties_and_theorems) but I am not sure this leads you anywhere.
To solve the differential equation, you'd better integrate $\frac{dx(t)}{\sin(x(t))}=dt$, giving $\ln(\tan(\frac{x(t)}{2}))-\ln(\tan(\frac{x_0}{2}))=t$ and $x(t)=2\ \arctan(\tan(\frac{x_0}{2})e^t)$