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?
 A: 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.
A: 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 $x =\frac {\pi} 3$.
A: $$\cos(2x)\cos(\pi/6-x)-\sin(2x)\sin(\pi/6-x)=0$$
$$\Rightarrow \cos(2x+\pi/6-x)=0$$ (using $\cos(a+b)=\cos a \cos b -\sin a \sin b$)
$$\Rightarrow x+\pi/6=\pi/2$$
$$\Rightarrow \boxed{x=\pi/3}$$
A: As an alternative, since $x=k\frac \pi 4$ and  $x = \frac{\pi}{6}+k\frac \pi 2$ are not solutions, we have that
$$\cos(2x)\cos\left(x - \frac{\pi}{6}\right) = \sin(2x)\sin\left(\frac{\pi}{6} - x\right) \iff \frac{\sin(2x)}{\cos(2x)}=\frac{\cos\left(\frac{\pi}{6} - x\right)}{\sin\left(\frac{\pi}{6} - x\right)}$$
$$\iff \tan (2x)=\cot\left(\frac{\pi}{6} - x\right)$$
and since
$$\tan A=\cot B \iff A=\frac \pi 2-(B+k\pi)$$
we obtain
$$2x=\frac \pi 2-\frac{\pi}{6} + x+k\pi \iff x=\frac \pi 3 +k\pi $$
