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?
 A: Set $p=\cos\beta$ and $q=\sin\beta$; then your equality can be written
$$
\cos\alpha+p-p\cos\alpha+q\sin\alpha=\frac{3}{2}
$$
or
$$
(1-p)\cos\alpha+q\sin\alpha=\frac{3}{2}-p
$$
Set $X=\cos\alpha$ and $Y=\sin\alpha$; then the equation becomes
$$
\begin{cases}
(1-p)X+qY=\frac{3}{2}-p\\[6px]
X^2+Y^2=1
\end{cases}
$$
and the distance of the line from the origin should be $\le1$, or the line and the circle wouldn't meet:
$$
\frac{(\frac{3}{2}-p)^2}{(1-p)^2+q^2}\le1
$$
that becomes
$$
\frac{9}{4}-3p+p^2\le 1-2p+p^2+q^2
$$
that is, recalling that $p^2+q^2=1$,
$$
p^2-p+\frac{1}{4}\le0
$$
Can you finish up?
A: 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 $0<\alpha<\pi$
