TITLE: There is a triangle, sides = a,b,c and its angle = $\alpha, \beta, \gamma$
QUESTION [0 upvotes]: $b=2\sqrt3$
$\sin(\alpha + \beta) + \cos(2\alpha-\beta) = 2$
Find a
I tried
$\frac{\sin\alpha}{a}=\frac{\sin\beta}{2\sqrt3}$
Or, 
Using identities 
$\sin(\alpha + \beta) = \sin\alpha\cdot\cos\beta + \cos\alpha\cdot\sin\beta$
But i can't solve it.
What formula should i use?

REPLY [0 votes]: Hint 
For $x,\sin x,\cos x\le1$
$\implies\sin(\alpha+\beta)+\cos(2\alpha-\beta)\le2$
So, we need $\sin(\alpha+\beta)=1=\cos(2\alpha-\beta)$
Also, $0<\alpha,\beta<\pi$
$\implies0<\alpha+\beta<\pi$
$0-\pi<2\alpha-\beta<2\pi-0\implies2\alpha-\beta$ has to be $0$
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