<s>[INST] Demonstrate this triangle! Help Give a triangle ABC with 
$$\sin{\left(\frac{3A}{2}\right)}+\sin{\left(\frac{3B}{2}\right)}=2\cos{\left(\frac{(A-B)}{2}\right)}$$
Demonstrate that triangle ABC is equilateral triangle!!
Thank all!
P/S: I'm sorry. Because I speak English not good!
 [/INST] $\sin{\left(\dfrac{3A}{2}\right)}+\sin{\left(\dfrac{3B}{2}\right)}=2\sin{\left(\dfrac{3A+3B}{4}\right)}\cos{\left(\dfrac{3A-3B}{4}\right)}$
let $x=\cos{\dfrac{A-B}{4}},p=\sin{\left(\dfrac{3A+3B}{4}\right)} \implies 1\ge x > \dfrac{\sqrt{2}}{2}, 1 \ge p >0 $,then we have:
$p(4x^3-3x)=2x^2-1$, it is trivial LHS< RHS when $\dfrac{\sqrt{3}}{2} >x > \dfrac{\sqrt{2}}{2}$, now we prove then $1 \ge x \ge \dfrac{\sqrt{3}}{2}$, LHS $\le$ RHS
LHS $\le 4x^3-3x \implies 4x^3-3x \le 2x^2-1 \iff 4x^3-3x - 2x^2+1\le 0 \iff (x-1)(4x^2+2x-1) \le 0$ 
it is true. so the "=" will hold when $p=1$ and $x=1$
so we have $\dfrac{3A+3B}{4}=\dfrac{\pi}{2},A=B \implies A=B=C=\dfrac{\pi}{3}$
QED
Edit: this can be more simple way:
WOLG, let $A \ge B $, 
$x=\dfrac{A-B}{2} \ge 0 ,y=\dfrac{3(A+B)}{4} < \dfrac{3\pi}{4}, x<\dfrac{3\pi}{4} $, now it becomes:
$\sin{y}\cos{\dfrac{3x}{2}} = \cos{x}$
$\cos{x}$ is mono decreasing function on [$0,\pi)$,  if $x \ge 0, \dfrac{3x}{2} \ge x, \cos{\dfrac{3x}{2}} \le \cos{x}, 1 \ge \sin{y} >0, \implies \sin{y} \cos{\dfrac{3x}{2}} \le \cos{x} \implies \sin{y}=1,\cos{\dfrac{3x}{2}}=\cos{x} \implies y=\dfrac{\pi}{2},x=0$
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