TITLE: Solutions for $\cos(\alpha)+\cos(\beta)-2\cos(\alpha+\beta)=0$ with a certain value range.
QUESTION [5 upvotes]: To proof 
How prove this equation has only one solution $\cos{(2x)}+\cos{x}\cdot\cos{(\sqrt{(\pi-3x)(\pi+x)}})=0$ 
I need first an analytically (not: numerically) proof for the following problem:
Be $\enspace\displaystyle 0<\beta<\frac{2\pi}{3}<\alpha<2\pi$ . 
Then it exists exactly $\,$ one $\,$ solution $\,(\alpha;\beta)\,$ for $\enspace \cos(\alpha)+\cos(\beta)-2\cos(\alpha+\beta)=0 $ ?
(The answers below show: No.) 
Known: $\enspace\displaystyle (\alpha_0;\beta_0):=\left(\pi;\arccos\left(\frac{1}{3}\right)\right)\enspace$ is a solution.

REPLY [2 votes]: Rewrite the equation as
$$
\cos (\alpha )+\cos (\beta )-2 \cos (\alpha ) \cos (\beta )+2 \sin (\alpha ) \sin (\beta )=0
$$
use the Weierstrass substitution for $\beta$ (or for $\alpha$), with $t=\tan(\beta/2)$
$$
\cos (\alpha )+\frac{1-t^2}{1+t^2}-\frac{2 \left(1-t^2\right) \cos (\alpha )}{1+t^2}+\frac{4 t \sin (\alpha )}{1+t^2}=\frac{t^2 (3 \cos (\alpha )-1)+4 t \sin (\alpha )+1-\cos (\alpha )}{1+t^2}=0
$$
Then solve the quadratic
$$
t^2 (3 \cos (\alpha )-1)+4 t \sin (\alpha )+1-\cos (\alpha )=0
$$
This is a graph of the solutions:

So there are infinite solutions, some particularly nice are
$$
(\pi,\arccos(1/3)),\quad(4\pi/3,\arccos(11/14)),\quad(3\pi/2,\arccos(2/\sqrt{5}))\quad(5\pi/3,\arccos(\sqrt{11/12}))
$$