Solve $\sin \frac{\alpha - \beta}2 + \sin \frac{\alpha - \gamma}2 + \sin \frac{3\alpha}2 =\frac 3 2$ Solve the following trigonometric eqation where $\alpha, \beta, \gamma$ are angles in a triangle ($\alpha + \beta + \gamma = 180$):
$$\sin \frac{\alpha - \beta}2 + \sin \frac{\alpha - \gamma}2 + \sin \frac{3\alpha}2 =\frac 3 2$$
Transforming it into $2 \sin \frac{3\alpha-180}4 \cos \frac{\beta - \gamma}4 + \sin \frac{3\alpha}2 =\frac{3}2$ 
and
$\cos \frac{\beta - \gamma}4 =\cos \frac{180- \alpha}4 + 2 \sin \frac{\beta}4 \sin \frac{\gamma}4$ is as far as I came.

When we exchange $\beta$ and $\gamma$ in the equation, we get the same equation back.  To me, that means $\beta = \gamma+4k\pi$, where $k$ is an integer, is a third equation that we can use.  Of course, only $k=0$ satisfies the constraints that $\alpha+\beta+\gamma=\pi$ and $\alpha, \gamma, \beta>0$, since they are angles in a triangle.  I haven't been able to derive it, but $\alpha=5\pi/9$, $\beta=2\pi/9$, and $\gamma=2\pi/9$ does appear to be the solution.
