Solving for $k$ in $\cos \frac{\pi}k - \cos \frac{2\pi}k = P$

Question

How do I solve trigonometric equations of the type $$\cos \frac{\pi}k - \cos \frac{2\pi}k = P$$ where P is a real constant within the range of the left side.
(By solving, I mean finding the value of k)

I could convert this into product of two $\sin$ functions but I don't think that will help. The value of P could indeed shed light on the value of k.
For example: If $P = \large\frac{\sqrt 3 - 1}2$; A little guess work and you get $k=6$. But this guess work almost instantly fails for $\large\frac{\sqrt 3 + 1}2$.
All help will be appreciated.

Answer 1

Guide:

$$\cos \frac{\pi}{k} - (2 \cos^2 \frac{\pi}{k} -1)=P $$

Hence this is a quadratic equation in $\cos \frac{\pi}k$

After we solve for the possible values of $\cos \frac{\pi}k $, we can find all possible values for $\frac{\pi}k$ and then solve for $k$.

Answer 2

$$\cos \frac{\pi}k - \cos \frac{2\pi}k = P$$ $$\cos \frac{\pi}k - 2\cos^2 \frac{\pi}k+1 = P$$ Substitute $y=\cos \frac{\pi}k $ $$y - 2y^2 = P-1$$ Solve the quadratic equation $$ 2y^2 -y+P-1 =0$$