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?

The equation is $\cos (2x) \cos (x-\frac {\pi} 6)+\sin (2x)\sin (x-\frac {\pi} 6)=0$. This is same as $\cos (2x-(x-\frac {\pi} 6))=0$ or $\cos (x+\frac {\pi} 6)=0$. So  $x+\frac {\pi} 6=\frac {(2n+1)\pi} 2$ for some integer $n$. For $x \in (0,\frac {\pi} 2)$ we must have $n=0$ so $x =\frac {\pi} 3$.
