Solving this second order differential equation $m\frac{d^2x}{dt^2}+c\frac{dx}{dt}+k\sin{x}=0$ I am investigating something in Physics and found out that I will have to solve this equation:
$$m\frac{d^2x}{dt^2}+c\frac{dx}{dt}+k\sin{x}=0$$
I haven't even learned how to solve ones that involve $kx$ and so I am very troubled by the $k\sin{x}$. Is this possible to solve? If so how could it be solved for $x$?

Edit: I can't assume that $\sin{x}\approx x$ since I am investigating cases in which such an assumption is not applicable (at large angles of $x$)
 A: I don't think there is a way to solve this one exactly. However, I can help you with the undamped case, that is $c=0$. In this case, using $\gamma=k/m$, the equation reads
$$y''(x)=-\gamma\sin{y(x)}$$
Which is a 1D sine-Gordon equation. Multiplying by $y'(x)$ and integrating, we get the energy function
$$E=\frac{(y'(x))^{2}}{2}-\gamma\cos{y(x)}$$ 
From this,
$$y'(x)=\pm\sqrt{2(E+\gamma\cos{y(x)})}$$
Which leads to
$$x-x_{0}=\pm\frac{1}{\sqrt{2E}}\int^{y(x)}\frac{dz}{\sqrt{1+\frac{\gamma}{E}\cos{z}}}=\pm\sqrt{\frac{2E}{E^{2}+\gamma^{2}}}\mathcal{F}\Big{(}\frac{y(x)}{2}\Big{|}\frac{2\gamma}{E+\gamma}\Big{)}$$
Where $\mathcal{F}(z|k)$ is the incomplete elliptic integral of the first kind. Now, we can use a "physicist's" style first order perturbation theory in $c$, to account for finite, but very small damping we assume $c\ll 1$. Let $y(x)=y_{0}(x)+cy_{1}(x)$
Then
$$\sin{y(x)}\approx\sin{y_{0}(x)}+cy_{1}(x)\cos{y_{0}(x)}$$
And the equations for the functions are
$$my_{0}''(x)+k\sin{y_{0}(x)}=0$$
$$my_{1}''(x)+y_{0}'(x)+ky_{1}(x)\cos{y_{0}(x)}=0$$
The solution to the first equation we already know. The second equation is linear second order ode, which no one knows how to solve, unfortunately. 
