If $0< \alpha, \beta< \pi$ and $\cos\alpha + \cos\beta-\cos (\alpha + \beta) =3/2$ then prove $\alpha = \beta= \pi/3$ 
If $0< \alpha, \beta< \pi$ and $\cos\alpha + \cos\beta-\cos (\alpha + \beta) =3/2$ then prove $\alpha = \beta= \pi/3$

How do I solve for $\alpha$ and $\beta$ when only one equation is given? By simplification I came up with something like 
$$
\sin\frac{\alpha}{2} \sin\frac{\beta}{2} \cos \frac {\alpha +\beta}{2}=\frac{1}{8}.
$$
I don't know if this helps. How to do this?

HINT:
Use Prosthaphaeresis Formula on $\cos\alpha,\cos\beta$
and Double angle formula on $$\cos 2\cdot\dfrac{\alpha+\beta}2$$ to get
$$2\cos^2\dfrac{\alpha+\beta}2-2\cos\dfrac{\alpha-\beta}2\cos\dfrac{\alpha+\beta}2+\dfrac12=0\ \  \ \ (1)$$ which is a Quadratic Equation in $\cos\dfrac{\alpha+\beta}2$  whose discriminant must be $\not<0$
i.e.,  $$\left(2\cos\dfrac{\alpha-\beta}2\right)^2-4=-2\sin^2\dfrac{\alpha-\beta}2\ge0$$
But for real $\dfrac{\alpha-\beta}2,$ $$\sin^2\dfrac{\alpha-\beta}2\ge0$$
So, we have $$\sin^2\dfrac{\alpha-\beta}2=0\iff\sin\dfrac{\alpha-\beta}2=0$$
$\implies\dfrac{\alpha-\beta}2=n\pi$ where $n$ is any integer
But as $0<\alpha,\beta<\pi.n=0\implies\alpha-\beta=0$
So, $(1)$ is reduced to $$0=2\cos^2\beta-2\cos\beta+\dfrac12=\dfrac{(2\cos\beta-1)^2}2$$
So, $\cos\beta=\dfrac12=\cos\dfrac\pi6\implies\beta=2m\pi\pm\dfrac\pi6$  where $m$ is any integer
But as $0<\alpha<\pi$
