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.
 A: I took an approach which may be only superficially different from abel's because of the underlying trig identities and follows a lead SoulEater was taking, but along a slightly different path.
We can get rid of $ \ \alpha \ $ , to start off, by writing $ \ \alpha - \beta \ = \ \pi - 2 \beta - \gamma \ $ and, similarly, $ \ \alpha - \gamma \ = \ \pi -  \beta - 2\gamma \ $ .  From this, we have
$$ \sin \left( \frac{\alpha - \beta}{2}  \right) \ = \ \sin \left( \frac{\pi}{2} - [ \ \beta + \frac{\gamma}{2} \ ] \right) \ = \ \cos \left( \ \beta + \frac{\gamma}{2} \ \right) \ \  $$
and $ \ \sin \left( \frac{\alpha - \gamma}{2}  \right) \ = \ \cos \left( \  \frac{\beta}{2} + \gamma \ \right) \   $ .  [So far, this looks like what abel did, while deriving the variable transformations.]  We will also write
$$ \sin \frac{3 \alpha}{2} \ = \ \sin \left( \frac{3}{2}  [ \pi - \beta - \gamma ] \right) \ = \ \sin \left( \frac{3 \pi}{2}  - \ \frac{3}{2} [\beta + \gamma ] \  \right) \ = \ -\cos \left( \frac{3}{2} [\beta + \gamma ] \  \right) \ \ . $$
The original equation is now  
$$ \cos \left( \ \beta + \frac{\gamma}{2} \ \right) \ + \ \cos \left( \  \frac{\beta}{2} + \gamma \ \right) \ - \ \cos \left( \frac{3}{2} [\beta + \gamma ] \  \right) \ = \ \frac{3}{2} \ \ . $$
Applying a "sum-to-product rule" on the first two terms here gives
$$ 2 \ \cos \left( \frac{3}{4} [\beta + \gamma ] \  \right)  \cos \left( \frac{1}{4} [\beta - \gamma ] \right)  - \ \cos \left( \frac{3}{2} [\beta + \gamma ] \  \right) \ = \ \frac{3}{2} \ \ . $$
This seemingly doesn't appear to have done much of anything useful, but it turns out we've actually finished the "hard part"...
We will label the angles  $ \ \Theta \ = \ \frac{3}{4} [\beta + \gamma ] \ $ and $ \ \Phi \ = \ \frac{1}{4} [\beta - \gamma ] \ $ , allowing us to write
$$ 2 \ \cos \Theta \   \cos \Phi \  - \ \cos \ (  2 \Theta   ) \ = \ \frac{3}{2} \ \ \Rightarrow \ \ 2 \ \cos \Theta \   \cos \Phi \  - \ 2 \ \cos^2 \Theta  \ + \ 1 \ = \ \frac{3}{2} $$
$$ \Rightarrow \ \  \cos \Theta \   \cos \Phi \  -  \ \cos^2 \Theta  \  = \ \frac{1}{4} \ \ . $$
Now, if $ \ \beta \ = \ \gamma \ $ , then $ \ \Phi \ = \ 0 \ $ and we just have the quadratic equation in $ \ \cos \Theta \ $ , 
$$     \cos^2 \Theta  \ - \ \cos \Theta \  + \ \frac{1}{4} \ = \ \left( \cos \Theta  \ - \  \frac{1}{2} \right)^2 \ = \ 0 \ \ . $$
Since we are limiting ourselves to angles of a triangle, we obtain
$$ \cos \left( \frac{3}{4} [\beta + \gamma ] \  \right) \ = \ \frac{1}{2} \ \ \Rightarrow \ \ \frac{3}{4} [\beta + \gamma ]  \ = \ \frac{\pi}{3} \ \ \Rightarrow \ \ \beta \ + \ \gamma   \ = \ \frac{4 \pi}{9} \ \ . $$
Thus,  $ \ \beta \ = \ \gamma \ = \ \frac{2 \pi}{9} \ \ \Rightarrow \ \ \alpha \ = \ \frac{5 \pi}{9} \ $ .
Are other solutions permissible?  If we allow $ \ \beta \ \neq \ \gamma \ $ and call $ \  \cos \Phi \ = \ \varphi \ $ , our quadratic equation becomes $ \     \cos^2 \Theta  \ - \ \varphi \ \cos \Theta \  + \ \frac{1}{4}  = \ 0 \  $ , which has potential solutions 
$$ \cos \Theta \ = \ \frac{\varphi \ \pm \ \sqrt{ \ \varphi^2 \ - \ 1 } } {2} \ \ . $$
But real solutions for $ \ \cos \Theta \ $ would only be possible here for $ \ \varphi \ = \ \cos \Phi \ \ge \ 1 \ $ .  So the only solution for our original equation is the one we've already found with $ \ \varphi \ = \ 1 \ $ or $ \ \beta \ = \ \gamma \ $ .
A: 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.
A: 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.
