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!
 A: If $A=60°$ and $B=60°$, the equation has a simple solution: $2=2$. So the triangle is equilateral
A: Let $x=A/2$ and $y=B/2$, and define $$f(x,y)=2 \cos(x-y)-\sin(3x)-\sin(3y),$$ so that you are looking for zeroes of $f$. Restrict $f$ to the domain $R=[0,\pi/2] \times [0,\pi/2].$ In this domain, all possible values of $A,B$ which can be angles in a triangle are represented, since e.g. $0<A<\pi$ implies $0<x<\pi/2.$ Other impossible combinations are included as well, but this domain is simple and includes all the relevant possibilities.
The function $f(x,y)$ must have an absolute maximum and an absolute minimum on the region $R$ since $f$ is continuous and $R$ is compact. Since $f(0,0)=2$ and $f(1,0) \approx 0.9394,$ we see the max is 2 or more (in fact it is 4 at $x=y=\pi/2$ which doesn't correspond to a triangle anyway, since then $A=B=180$ degrees). Our claim now is that the minimum of $f$ is $0.$
Now the functions being smooth, we may (after enlarging $R$ a little backward and forward on each axis) assume that at any minimal pair $(x,y)$ the two partials of $f$ both vanish.
When these partials are computed and added, we obtain (since $f_x=0,f_y=0$):
$$f_x+f_y=-3\cos(3x)-3\cos(3y)=0.$$
This means that either $3x=3y$ and $\cos(3x)=0$, which leads back to the maximum at $(\pi/2,\pi/2)$, or $3y=\pi \pm 3x$. Taking the plus sign here gives $y=\pi/3+x$ for which $f(x,\pi/3+x)=1$, so no solution comes from the $+$ choice.
The remaining choice is $y=\pi/3-x$, and for this we have
$$f(x,\pi/3-x)=-2\sin(3x)+2\sin(2x+\pi/6).$$
This must be zero at a critical point, which gives either $3x=2x+\pi/6$ or else $3x=\pi-(2x+\pi/6)$, both of which have the solution $x=\pi/6$. Thus the critical point must be at $(\pi/6,\pi/3-\pi/6)=(\pi/6,\pi/6)$, and since $A=2x,B=2y$ we have at the minimal value (which is in fact $0$) that $A=B=\pi/3$, i.e. in degrees they are both 60, so that the third angle is also 60, finally giving an equilateral triangle.
A: $\sin{\left(\dfrac{3A}{2}\right)}+\sin{\left(\dfrac{3B}{2}\right)}=2\sin{\left(\dfrac{3A+3B}{4}\right)}\cos{\left(\dfrac{3A-3B}{4}\right)}$
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 $x \ge 0, \dfrac{3x}{2} \ge x, \cos{\dfrac{3x}{2}} \le \cos{x}, 1 \ge \sin{y} >0, \implies \sin{y} \cos{\dfrac{3x}{2}} \le \cos{x} \implies \sin{y}=1,\cos{\dfrac{3x}{2}}=\cos{x} \implies y=\dfrac{\pi}{2},x=0$
