$y=\tan x-\dfrac{1-\cos3x}{1-\sin3x}$
Để $\tan x$ xác định: $x\ne \dfrac{\pi}{2}+k\pi$
ĐK: $\sin3x\ne 1$
$\Leftrightarrow 3x=\dfrac{\pi}{2}+k2\pi$
$\Leftrightarrow x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}$
$\to D=\mathbb{R}$ \ $\{\dfrac{\pi}{2}+k\pi; \dfrac{\pi}{6}+\dfrac{k2\pi}{3}\}$