Giải thích các bước giải:
$\cot x.\cot(2x-\dfrac{\pi}{6})+1=0$
$\to \dfrac{\cos x}{\sin x}.\dfrac{\cos(2x-\dfrac{\pi}{6})}{\sin(2x-\dfrac{\pi}{6})}+1=0$
$\to \cos x.\cos(2x-\dfrac{\pi}{6})+\sin x.\sin(2x-\dfrac{\pi}{6})=0$
$\to \cos (\dfrac{-\pi+12x}{6}-x)=0$
$\to x=\dfrac{2\pi}{3}+2k\pi, x=\dfrac{5\pi}{3}+k2\pi$