How to solve $\frac{d^2x}{dt^2}=k\sin(2x)$ How to solve $$\frac{d^2x}{dt^2}=k\sin(2x)$$
 A: $$\frac{d^2x}{dt^2}=-k \sin 2x$$
Multiply by $2 \frac{dx}{dt}$ to write
$$2 \frac{dx}{dt}\frac{d^2x}{dt^2} =2k \frac{dx}{dt} \sin 2x \implies \frac{d}{dt} \left(\frac{dx}{dt}\right)^2=2\frac{dx}{dt}k\sin 2x$$
Integrate.w.r.t. $t$ both sides
$$\int d \left(\frac{dx}{dt}\right)^2 =\int 2k \sin 2x \frac{dx}{dt} dt$$
$$\implies \left(\frac{dx}{dt}\right)^2=-k \cos 2x+A ~~~(1)$$
$$\implies \frac{dx}{dt} = \pm \sqrt{A-k \cos 2x} \implies \int \frac{dx}{\sqrt{A-k\cos 2x}}=\pm \int dt +B$$
Ny knowing $dx/dt$ at some time $t=t_0$ we can fix $A$ and the $\pm$ sign. Similarly, knowing $x(t_0)$ we can fix $B$.
However the above integral cannot be done in terms of simple functions. This integral is evaluated in terms of  ELLIPTIC integrals.
You may see
https://en.wikipedia.org/wiki/Elliptic_integral
A: If you switch variables, the equation write
$$\frac{t''(x)}{[t'(x)]^3}=-k \sin(2x)$$ Then reduction of order $p=t'(x)$ leads to
$$t'(x)=\frac{1}{\sqrt{ c_1-k \cos (2 x)}}$$ leading to the elliptic integrals @Yves Daoust already mentioned in comments.
Otherwise, working with $x(t)$, I suppose that you could face the amplitude for Jacobi elliptic functions.
