Say I have an equation of the form: $$ y = A + B\sin (x) +C\sin (x+Dy) $$ on the domain $0<x<2\pi$.
I want to get $y$ as a function solely of $x$, i.e. remove the $y$-dependence from the R.H.S. This cannot be done analytically, but are there any useful tricks to analytically obtain approximate solutions? The trouble is that $x$ and $y$ can take on arbitrary values, i.e. I can't take them to be small and make a first-order approximation of $\sin$, for example.
On
Start expanding the last sine and use the tangent half-angle substitution $x=2 \tan ^{-1}(t)$ to mka a quadratic equation in $t$ $$(a+c \sin (d y)-y)+2 (b+c \cos (d y))t+ (a-c \sin (d y)-y)t^2=0$$ which will show real solutions if $$(b^2+c^2-a^2)+2 a y+2 b c \cos (d y)-y^2 >0$$ If this is the case, you have two solutions $t_\pm$ and $x_\pm=2 \tan ^{-1}(t_\pm)$.
There is no way to get $y(x)$.
You can get $x$ solely in terms of $y$ $$y= A + B\sin(x) + C\sin(x+Dy)$$ $$y = A + B\sin(x) + C\sin(x)\cos(Dy) + C\sqrt{1-\sin^2 (x)}\sin(Dy)$$ $$z=\sin(x)$$ $$y = A + Bz + Cz\cos(Dy) + C\sqrt{1-z^2}\sin(Dy)$$ $$w=\cos(Dy)$$ $$y= A + Bz + Czw + C\sqrt{1-z^2}\sqrt{1-w^2}$$ Which yields: $$z_{\pm}=\frac{\pm\sqrt{-C^2 \left(w^2-1\right) \left(-A^2+2 A y+B^2+2 B C w+C^2-y^2\right)}-A (B+C w)+B y+C w y}{B^2+2 B C w+C^2}$$