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.

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 $\theta$,
$$\theta_\pm=2 \tan^{-1}(t_\pm)$$
