# Solution for following trigonometric equation?

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.

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If you make the substitution $\theta=2\arctan\,u$, you get an quartic equation in terms of $u$. –  Ｊ. Ｍ. Dec 5 '11 at 16:40
@J.M. : Okay. I will get back to you on that one :) –  Ole Thomsen Buus Dec 5 '11 at 17:04

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 $\theta$.

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Thanks! I am getting there myself eventually, but you are faster than me :) All three parameters $x$, $y$, and $r$ will always be equal to or above zero. I guess $G_\omega$ is always equal to or below zero. One question: Is your choice of variable $t$ identical to your use of $u$ in your comment above? I ask because $\theta = \omega t$, and since we know $\omega$ we could solve it directly for time $t$ now. But that is not what you have done since $G_\omega$ remains i guess? Could we (you) change $t$ to $u$? Thanks. I can solve it for time myself. Thanks! –  Ole Thomsen Buus Dec 5 '11 at 19:56
Yes, I just used $t$ here instead of $u$ in my comment. If you are already using $t$ for something different, you should replace all instances of it here with some other variable. –  Ｊ. Ｍ. Dec 6 '11 at 13:58
Your derivation into a quartic has been confirmed. –  Ole Thomsen Buus Dec 7 '11 at 17:29
Good to know. :) –  Ｊ. Ｍ. Dec 7 '11 at 17:31
I am working with getting a nice general solution for the roots for $u$. It seems it is a quasi-symmetric quartic: cogitosmath.wordpress.com/2010/02/22/quasi-symmetric-quartic. I created a new question that deals with finding the roots: math.stackexchange.com/questions/89312/…. I tried in Mathematica with a numerical example, and there seems to be two real and two complex roots. If $x^2+y^2 > r^2$ there will be four complex roots ($x$ and $y$ must be points inside a circle with radius $r$). –  Ole Thomsen Buus Dec 7 '11 at 22:13