I want to solve the system for $x$, $y$ and $z$. Is there any smart trick to solve it?
$$\begin{cases} 2a(ax+by)+2c(cx+dy)+2zx=0 \\ 2b(ax+by)+2d(cx+dy)+2zy=0 \\ x^2+y^2-1=0\end{cases}$$
$a,b,c,d \in \mathbb R$ and $(x,y,z)\in \mathbb R^3$. It is a part of another problem I am working on.
Make the substitutions
$$ a=R_1 \cos(\phi_1), b=R_1\sin(\phi_1) \\ c=R_2 \cos(\phi_2), d=R_2\sin(\phi_2) \\ x=\cos (\theta), y=\sin(\theta) $$ Eliminate $z$ to get an equation in $\theta$ $$\sin \left(3\,\left(\theta-{ \phi_2}\right)\right)\, { R_2}^3+\sin \left(\theta-{ \phi_2}\right)\,{ R_2}^3+ \sin \left(3\,\left(\theta-{ \phi_1}\right)\right)\,{ R_1}^3 +\sin \left(\theta-{ \phi_1}\right)\,{ R_1}^3=0$$