I have the following trigonometric equation in $\theta$:
$$0=G_{\omega}(1/r^2)({\csc^2}\theta){(r\cos\theta-x)}^2+(\cot\theta)(r\cos\theta-x)+r\sin\theta-y.$$
Is there an analytical solution for $\theta$ to this equation for known values of $G_w \in \mathbb{R}$, $x \in \mathbb{R}$, and $y \in \mathbb{R}$. By the way: $G_w = -(g/2){\omega}^{-2}$. The above equation describes the parabolic path of the free fall trajectory of a particle in $(x,y)$. It has the following initial position and velocity components: $x_0=r\omega\cos\theta$, $y_0=r\omega\sin\theta$, $\dot{x}_0 = -r\omega\sin\theta$, and $\dot{y}_0=rw\cos\theta$ (where $\theta=\omega t$; the particle is being thrown of the inner wall of a cylinder with angular velocity $\omega$ and radius $r$).
Numerical methods can be used, and have been used. But I was looking for an analytical solution.
Thanks.
Making the Weierstrass substitution $\theta=2\arctan\,t$ and rearranging yields the quartic equation:
$$G_\omega(r+x)^2 t^4+2r^2 t^3 (r+x)+(2G_\omega (x^2-r^2)-4 r^2 y)t^2+2r^2 (r-x)t+G_\omega (r-x)^2=0$$
You haven't given any other restrictions on your parameters except for them being real, so it's up to you to pick out which of this polynomial's four roots is real. From the real roots, you can easily obtain the needed $\theta$.