<s>[INST] Solving awkward quadratic equation to obtain "nice" solution. 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.
 [/INST] 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})} $
</s>