Solving $\frac{a\cos(\phi-\xi)}{1+d\cos(\phi)}=-\frac{b\cos(\theta-\xi)}{1-d\cos(\theta)}$ for $\theta$ $$\frac{a\cos(\phi-\xi)}{1+d\cos(\phi)}=-\frac{b\cos(\theta-\xi)}{1-d\cos(\theta)}$$ Solve for $\theta$. With $0\le\phi<2 \pi,$ $0\le\theta<2 \pi,$ $0\le \xi<\pi,$ $0 \le d<1$ and $a,b\ge 1.$
I am able to solve for $\theta$ only when either the denominator of R.H.S. is unity or when $\xi=0$.
In case an exact solution is not possible, ways to find an approximate solution is appreciated. If this is a trivial problem to solve, please excuse, I am not a Mathematician. Thank you for your time.
 A: First, rewrite the equation as
$$-\frac a b\frac{ \cos(\phi-\xi)}{1+d  \cos(\phi)}=\frac{  \cos(\theta-\xi)}{1-d  \cos(\theta)}$$ The lhs is a constant : call it $k$. So, the equation is now
$$k (1-d  \cos(\theta)=\cos(\theta-\xi)$$ Expand the cosine to make it
$$k (1-d  \cos(\theta))=\sin (\theta ) \sin (\xi )+\cos (\theta ) \cos (\xi )$$ Now, use the tangent half-angle substitution $t=\tan \left(\frac{\theta }{2}\right)$, simplify and arrive to the quadratic equation
$$\big[(d+1)k+\cos (\xi )\big]\,t^2-2  \sin (\xi )\,t-\big[(d-1) k+\cos (\xi )\big]=0$$ which will or not have one or two real solutions.
Solve it for $t_\pm$ and, basc to $\theta$,
$$\theta_\pm=2 \tan^{-1}(t_\pm)$$
A: Assuming that $\theta$ is the only unknown, with the other quantities given, write $$\lambda=-\frac1b\frac{a\cos(\phi-\xi)}{1+d\cos\phi}$$ and now solve $$\frac{\cos(\theta-\xi)}{1-d\cos\theta}=\lambda$$
$$\implies \lambda-\lambda d\cos\theta=\cos\theta\cos\xi+\sin\theta\sin\xi$$
$$\implies \sin\theta\sin\xi+\cos\theta[\cos\xi+\lambda d]=\lambda$$
Now write the LHS as $$R\sin(\theta+\epsilon)=R\sin\theta\cos\epsilon+R\cos\theta\sin\epsilon$$
So that $$\sin\xi=R\cos\epsilon$$ and $$\cos\xi+\lambda d=R\sin\epsilon$$
Then $$\tan\epsilon=\frac{cos\xi+\lambda d}{\sin\xi}$$
And $$R^2=\sin^2\xi+\cos^2\xi+2\lambda d\cos\xi+\lambda^2d^2=1+2\lambda d\cos\xi+\lambda^2d^2$$
So now we just have to solve $$R\sin(\theta+\epsilon)=\lambda\implies\sin(\theta+\epsilon)=\frac{\lambda}{R}$$
So if the principal value from this is $\alpha$, then the solution for $\theta$ is given by
$$\theta+\epsilon=\begin{cases}\alpha\\\pi-\alpha\end{cases}+n\cdot2\pi, n\in\mathbb{Z}$$
