Đáp án:
$ x=\dfrac{\pi}{36}+\dfrac{k\pi}{6}$
Giải thích các bước giải:
$cos6x-\sqrt[]{3}sin6x=0$
$\rightarrow\dfrac{1}{2}cos6x-\dfrac{\sqrt[]{3}}{2}sin6x=0$
$\rightarrow cos6x.cos\dfrac{\pi}{3}-sin6x.sin\dfrac{\pi}{3}=0$
$\rightarrow cos(6x+\dfrac{\pi}{3})=0$
$\rightarrow 6x+\dfrac{\pi}{3}=\dfrac{\pi}{2}+k\pi$
$\rightarrow x=\dfrac{\pi}{36}+\dfrac{k\pi}{6}$