Solving awkward quadratic equation to obtain "nice" solution.

Question

I would like to solve the following quadratic equation to get a "nice" analytic solution for $\rho$.

$\rho^2(r\sin\theta-2nr^2)+\rho(2nr^3-2r^2\sin\theta-2\sin\theta+2nr)-2nr^2+3r\sin\theta=0$

where $r=1-\cos\theta$ (I cannot see how this could be used to simplify the quadratic equation.)

I would hope that the solution to be found which be of a simple form as that would correspond nicely to the larger problem I am working on.

Also $\theta=\pi/n$

EDIT: I would hope that the solution $\displaystyle \rho=1-\frac{sin(\frac{\pi}{n})}{n}$ would be one of the solutions to the quadratic.

Answer 1

It is not a particularly simple form.

Solution:

$\rho = (R \pm S)/Q$

Where

$Q = \textrm{ sin}( \frac{\pi}{2 n} )^3( 32 n - 16 \textrm{ cot}(\frac{\pi}{2 n}))$

$R = -12 \textrm{ cos}(\frac{\pi}{2 n}) + 6 \textrm{ cos}(\frac{3 \pi}{2 n} ) - 2 \textrm{ cos}( \frac{5 \pi}{2 n}) + 2 n \left(14 \textrm{ sin}( \frac{\pi}{2 n}) - 5 \textrm{ sin}( \frac{3\pi}{2 n}) + \textrm{ sin}( \frac{5\pi}{2 n}) \right)$

$S = \sqrt{2} \sqrt{22 - 58 n^2 + 6 (1 + 17 n^2) \textrm{ cos}( \frac{\pi}{n} ) - 56 n^2 \textrm{ cos}(\frac{2\pi}{n}) + (9 + 11 n^2) \textrm{ cos}(\frac{3\pi}{n}) + 2 (-3 + n^2) \textrm{ cos}(\frac{4\pi}{n}) - (-1 + n^2) \textrm{ cos}(\frac{5\pi}{n}) - 20 n \textrm{ sin}( \frac{\pi}{n}) + 8 n \textrm{ sin}( \frac{2\pi}{n}) - 22 n \textrm{ sin}(\frac{3\pi}{n}) + 12 n \textrm{ sin}( \frac{4\pi}{n} ) - 2 n \textrm{ sin}( \frac{5\pi}{n})} $

Answer 2

I don't think that it is so easy as to solve it, it would be possible if we had the expression leading to this which is relatively simple and we could get simple closed expression,Anyways if you're asking for a solution:

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Solving in wolfram alpha at here I got the following two roots:

Let $$\cos(\lambda\pi/n)=c_{\lambda}$$ And similiarly $$\sin(\lambda\pi/n)=s_{\lambda}$$ Now let $$\color{red}{S_Q}=62n^2-106n^2c_1+72n^2c_2-37n^2c_3\\+10n^2c_4-n^2c_5-36ns_1+32ns_2-38ns_3\\+16ns_4-2ns_5+6c_1+9c_3+c_5+22$$ And $$\color{red}{D_N}=-12ns_1+4ns_3+c_1-c_3$$ And $$\color{red}{R_D}=-14ns_1+5ns_3-ns_5+6c_1-3c_3+c_5$$ Then roots are: $$\rho_1=\frac{\pm1}{\sqrt{2}\color{red}{D_N}}[\sqrt{\color{red}{S_Q}}-\color{red}{R_D}]$$