b) tập xác định là R vì [TEX]-1\geqslant sinx \geqslant 1[/TEX]
a) [tex]\dpi{120} cos(3x-\frac{2\pi}{9})\neq 1=>3x-\frac{2\pi}{9}\neq k2\pi=>x\neq \frac{k2\pi+\frac{2\pi}{9}}{3}[/tex]
b) [tex]\dpi{120} \left\{\begin{matrix} cosx\neq 0=>x\neq k2\pi & \\ tan^2x-2\sqrt{3}tanx+3\neq 0<=>(tanx-\sqrt{3})^2\neq 0=>tanx\neq \sqrt{3} => x\neq \frac{\pi}{3}+k\pi& \end{matrix}\right.[/tex]