<s>[INST] 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.
 [/INST] i think the answer is $\alpha = 100^\circ, \beta = 40^\circ$ and $\gamma = 40^\circ$
here is my attempt at the solution. i will use a change of variable 
$$b = \beta+{1 \over 2}\gamma,\  c = \gamma + {1 \over 2}\beta$$ and the following inverse relations 
$$\alpha = 180^\circ - {2 \over 3}(b+c), \ \beta = {4 \over 3}(b - {c \over 2})\ , \gamma = {4 \over 3}(c - {b \over 2}) $$ this change of variable transforms 
$\sin({\alpha - \beta \over 2}) +  \sin({\alpha - \gamma \over 2}) + \sin{3\alpha \over 2} = {3 \over 2}$ into a symmetric $2\pi$-periodic equation 
$$f(b,c)=\cos b + \cos c - \cos(b+c) = {3 \over 2}.$$ observe that $f = 1$ on the boundary and $f(b,b) = 2\cos b - \cos 2b = -2\cos^2 b + 2 \cos b +1$ is maximum $3/2$
at $b = \pi/3 \text{ and } 5\pi/3.$ 
we will show that the global maximum of $f(b,c)$ on the square $[0,2\pi] \times [0,2\pi]$ is ${3 \over 2}$ at $b = c = \pi/3.$ 
at a local extremum, the partial derivatives 
$$f_b = -\sin b + \sin(b+c) = 0,\  f_c = -\sin c + \sin(b+c) = 0$$ 
so we need to solve $$\sin b = \sin c = \sin(b+c).$$
the solutions are $$c = b, c = \pi - b, c = 3\pi - b $$
only the case $c = b$ is relevant. we have $\sin b = \sin 2b$ an the solution is 
$\cos b = 1/2, \sin b = 0 $
now we can transform the $$b = c = 60^\circ$$ to the values for $\alpha, \beta$ and $\gamma$ claimed at the top of the post.
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