#### Answer

The solution of the given trigonometric equation is $\theta =0,\pi$

#### Work Step by Step

We have the equation as $\sin \theta =\tan \theta $.
By solving the equation,
$\begin{align}
& \sin \theta =\tan \theta \\
& \sin \theta =\frac{\sin \theta }{\cos \theta } \\
& \sin \theta =\frac{\sin \theta }{\sqrt{1-{{\sin }^{2}}\theta }} \\
& \sin \theta \times \sqrt{1-{{\sin }^{2}}\theta }=\sin \theta
\end{align}$
Then,
$\begin{align}
& \sin \theta \times \sqrt{1-{{\sin }^{2}}\theta }-\sin \theta =0 \\
& \sin \theta \left( \sqrt{1-{{\sin }^{2}}\theta }-1 \right)=0
\end{align}$
And from the two factors, we get
$\begin{align}
& \sin \theta =0 \\
& \theta =n\pi
\end{align}$
and,
$\begin{align}
& 1-{{\sin }^{2}}\theta =1 \\
& {{\sin }^{2}}\theta =0 \\
& \sin \theta =0 \\
& \theta =n\pi
\end{align}$
Therefore, we get $\theta =n\pi $.
The provided range is $\left[ 0,2\pi \right)$. Hence, $\theta =0,\pi $ are the solutions of the provided equation.