Answer
The solution in the interval $[0,2\pi )$ is $\frac{2\pi }{3}$.
Work Step by Step
We know that the period of the tangent function is $\pi $. In the interval $(0,\,\,\pi ]$, the only value for which the tangent function equals $\sqrt{3}$ is $\frac{\pi }{3}$.
All the solutions to $\tan \frac{x}{2}=\sqrt{3}$ are given by:
$\begin{align}
& \frac{x}{2}=\frac{\pi }{3}+n\pi \\
& x=\frac{2\pi }{3}+2n\pi
\end{align}$
Where, n is any integer. The solutions in the interval $[0,2\pi )$ are obtained by letting $n=0$. And the equation is calculated by taking first $n$ as 0. It can be further simplified as follows.
$\begin{align}
& x=\frac{2\pi }{3}+2n\pi \\
& =\frac{2\pi }{3}+\frac{0\times \pi }{3} \\
& =\frac{2\pi }{3}+0 \\
& =\frac{2\pi }{3}
\end{align}$
