Solve $\theta''+g\sin(\theta)=0$ I encountered the following differential equation when I tried to derive the equation of motion of a simple pendulum:
$\frac{\mathrm d^2 \theta}{\mathrm dt^2}+g\sin\theta=0$
How can I solve the above equation?

replacing $\sin\theta$ by $\theta$ (physically assuming small angle deflection) gives you a homogeneous second order linear differential equation with constant coefficients, whose general solution can be found in most introductory diff eq texts (or a google search).  this new equation represents a simple harmonic oscillator (acceleration proportional to displacement, like a spring force).
$$
\theta''+g\theta=0
$$
has solutions $A\cos(\sqrt{g}t)+B\sin(\sqrt{g}t)$.
so, for example, if the initial displacement is $\theta_0$ and initial angular velocity is $0$ then the solution is
$$
\theta_0\cos(\sqrt{g}t)
$$
