Answer
$$x=\frac{\pi }{6}, \qquad x=\frac{5\pi }{6}, \\ x= \frac{7\pi }{6}, \qquad x=\frac{11\pi }{6}$$
Work Step by Step
We solve as follows:
$$9\tan ^2 x -3=0 \quad \Rightarrow \quad \tan ^2 x = \frac{1}{3}\quad \Rightarrow \quad \sin x = \pm \frac{\sqrt{3}}{3} \quad \Rightarrow \quad x=\frac{\pi }{6}+ n\pi, \quad \text{ or } \quad x=\frac{5\pi }{6} + n \pi \quad n \in \mathbb{Z}$$So, $x= \frac{\pi }{6}$, $x= \frac{5\pi }{6}$, $x= \frac{7\pi }{6}$, and $x=\frac{11\pi }{6}$ are the only solutions in the interval $[0, 2\pi )$.
