<s>[INST] 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?
 [/INST] 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
$$2x=\frac \pi 2-\frac{\pi}{6} + x+k\pi \iff x=\frac \pi 3 +k\pi $$
</s>