Solving $\cos(2x)\cos\left(x - \frac{\pi}{6}\right) = \sin(2x)\sin\left(\frac{\pi}{6} - x\right)$ for $x\in(0,\pi/2)$ 
Solve this equation for $x\in (0 , \frac{\pi}{2})$
$$\cos(2x)\cos\left(x - \frac{\pi}{6}\right) = \sin(2x)\sin\left(\frac{\pi}{6} - x\right)$$

I gave a try using the $\cos(a-b)$ and $\sin(a-b)$ formulas, but it seems the problem complicated a little bit more.
Is any other elegant solution for this?

It is not complicated, the equation become
$$\cos(2x)\cos(x-\frac{\pi}{6})+\sin(2x)\sin(x-\frac{\pi}{6})=0$$
that is equivalent to
$$\cos(x+\frac{\pi}{6})=0$$
Thus the solution is
$x+\frac{\pi}{6}=\frac{\pi}{2}+k\pi$
$x=\frac{\pi}{3}+k\pi$
This is the most elegant solution for me.
