In physics I never solve the equation $\ddot\theta = \sin(\theta)$. Instead, we used the approximation $\theta = \sin(\theta)$ for small angles and then it was easy to solve. I didn't do any physics since a while but I was interested by the original equation. I tried to solve it with formal series but it was quite ugly and I give up quickly. But maybe it's well known so I'm asking the question here :
What are the solutions to the equation $\ddot\theta = \sin(\theta)$ ?
A first good step is multiplying by $2\dot{\theta}$, to get
$$ \frac{d}{dt} \dot{\theta}^2=2\sin \theta \dot{\theta} $$
which can be integrated to
$$\dot{\theta}^2=-2\cos \theta+A,$$
where $A$ is a constant. Thus
$$\frac{d \theta}{ \sqrt{A-2\cos \theta}}= \pm dt $$
and a second integration gives
$$F_A(\theta)=\pm(t-t_0) $$
where $F_A$ is a primitive function of $\frac{1}{\sqrt{A-2\cos \theta}}$ and $t_0$ is another constant.
EDIT: The solution given above is implicit for $\theta(t)$. Mathematica gives
$$\theta(t)= \pm 2 \text{am}\left(\frac{1}{2} \sqrt{\left(c_1-2\right) \left(t+c_2\right){}^2},-\frac{4}{c_1-2}\right) $$
where am is the Jacobi Amplitude function.