Solving pair of differential equations involving two functions $$\frac{d^2h(x)}{dx^2} = f(x)\sin(h(x))$$ 
$$\frac{d^2f(x)}{dx^2} = \cos(h(x))$$
Is it even possible to solve the equations for $f(x)$ and $h(x)$? If so, how would one solve them?
 A: I may give an idea, use an Ansatz that $h(x)$ is of form $h(x)= F(x) \sin(h(x))$, then $$h = F(x) \sin(h(x)) \implies h' = \frac{F' \sin(h)}{1-F\cos(h)} \implies h'' = \frac{ (F'' \sin(h) + F h' \cos(h))(1- F \cos(h)) - (F' \sin(h) (Fh'\sin(h) - F' \cos(h))}{(1-F\cos(h))^{2}}$$ so we must have
$$ h''(x)=  \frac{ (1- F \cos(h))F'' \sin(h) + F F' \sin(h)\cos(h) - F' \sin(h) \left(\frac{FF'\sin^{2}(h)}{1-F \cos(h)} - F' \cos(h)\right)}{(1-F\cos(h))^{2}} =\underbrace{\left[\frac{ (1- F \cos(h))F'' + F F' \cos(h) - F' \left(\frac{FF'\sin^{2}(h)}{1-F \cos(h)} - F' \cos(h)\right)}{(1-F\cos(h))^{2}} 
\right]}_{f(x)}\sin(h) $$
Now find (or maybe setup, is more appropriate) $F$ that satisfy $f''(x) = \cos(h(x)).$
A: $$\begin{cases}
\frac{d^2h(x)}{dx^2} = f(x)\sin(h(x))\\ 
\frac{d^2f(x)}{dx^2} = \cos(h(x))
\end{cases}$$
From the first equation : $\quad f(x)=\frac{\frac{d^2h(x)}{dx^2}}{\sin(h(x))}$
$\frac{df(x)}{dx} =\frac{\frac{d^3h(x)}{dx^3}}{\sin(h(x))}
-\frac{\frac{d^2h(x)}{dx^2}\frac{dh(x)}{dx}\cos(h(x))}{\sin^2(h(x))}$
$\frac{d^2f(x)}{dx^2} =\frac{\frac{d^4h(x)}{dx^4}}{\sin(h(x))}
-2\frac{\frac{d^3h(x)}{dx^3}\frac{dh(x)}{dx}\cos(h(x))}{\sin^2(h(x))}
-\frac{\left(\frac{d^2h(x)}{dx^2}\right)^2\cos(h(x))}{\sin^2(h(x))}
+\frac{\frac{d^2h(x)}{dx^2}\left(\frac{dh(x)}{dx}\right)^2}{\sin(h(x))}
+2\frac{\frac{d^2h(x)}{dx^2}\left(\frac{dh(x)}{dx}\right)^2\cos^2(h(x))}{\sin^3(h(x))}$
$$\cos(h)=\frac{h''''}{\sin(h)}
-2\frac{h'''h'\cos(h)}{\sin^2(h)}
-\frac{\left(h''\right)^2\cos(h)}{\sin^2(h)}
-\frac{h''\left(h'\right)^2}{\sin(h)}
+2\frac{h''\left(h'\right)^2}{\sin^3(h)}$$
The most likely, such a fourth order non-linear ODE of autonomous kind cannot be solved for a closed form of $h(x)$. So, I am afraid that the problem has to be solved only thanks to numerical calculus.
