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

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.

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 $x =\frac {\pi} 3$.

$$\cos(2x)\cos(\pi/6-x)-\sin(2x)\sin(\pi/6-x)=0$$ $$\Rightarrow \cos(2x+\pi/6-x)=0$$ (using $\cos(a+b)=\cos a \cos b -\sin a \sin b$) $$\Rightarrow x+\pi/6=\pi/2$$ $$\Rightarrow \boxed{x=\pi/3}$$