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?

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?
