Solutions for $\cos(\alpha)+\cos(\beta)-2\cos(\alpha+\beta)=0$ with a certain value range. 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. 
 A: 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
$$
(\pi,\arccos(1/3)),\quad(4\pi/3,\arccos(11/14)),\quad(3\pi/2,\arccos(2/\sqrt{5}))\quad(5\pi/3,\arccos(\sqrt{11/12}))
$$
A: It is not true that $(\alpha_0,\beta_0)=(\pi,\arccos(\frac{1}{3}))$ is the only solution. Consider for example $\alpha_1=\frac{5}{6}\pi$ and $g(x)=\cos(\alpha_1)+\cos(x)-2\cos(\alpha_1+x)$. We have
$$
g(\frac{\pi}{2})=
\cos(\frac{5}{6}\pi)+\cos(\frac{\pi}{2})-2\cos(\frac{4\pi}{3})=
-\frac{\sqrt{3}}{2}+0-2(\frac{-1}{2})=\frac{2-\sqrt{3}}{2} >0. \tag{1}
$$
Also, $$g(\frac{5\pi}{8})=
\cos(\frac{5}{6}\pi)+\cos(\frac{5\pi}{8})-2\cos(\frac{35\pi}{24})
\leq -\frac{\sqrt{3}}{2}-\frac{1}{3}+1=\frac{4-\sqrt{27}}{6}<0. \tag{2} $$
To justify (2), we need (a) $\cos(\frac{5\pi}{8}) \lt -\frac{1}{3}$ and (b) 
$\cos(\frac{35\pi}{24}) \gt -\frac{1}{2}$. The easiest one is (b) : it follows from
$\frac{35\pi}{24}\in (\frac{4\pi}{3},2\pi)$ and the fact that $\cos$ is decreasing on that interval. Regarding (a), note first that $\cos(\frac{5\pi}{8})=-\sin(\frac{\pi}{8})$.
Next, using Taylor's formula with Lagrange remainder, we have
$$
f(x)-f(0)-f'(0)x-f''(0)\frac{x^2}{2}-f'''(0)\frac{x^3}{6}=\frac{f^{(4)}(\xi)}{24}x^{4}
$$
for some $\xi\in[0,x]$. Applying this to $f(x)=\sin(x)$, we deduce $\sin(x)\geq x-\frac{x^3}{6}$ for $x\in [0,\frac{\pi}{2}]$. In particular,
$\sin(\frac{\pi}{8}) \geq h(\frac{\pi}{8})$. Since $h'(x)=1-2x^2$, we see that $h$
is increasing on $[0,\frac{1}{2}]$, so $h(\frac{\pi}{8})\geq h(\frac{3}{8})=\frac{375}{1024} \gt \frac{1}{3}$. This finishes the proof. 
