Answer
No solution in the interval $[0,2\pi )$.
Work Step by Step
We know that the period of the sine function is $2\pi $. In the interval $(0,\,\,\pi ]$ the only value for which the sine function is $-1$ is $\frac{3\pi }{2}$.
Therefore, all the solutions to $\sin \frac{2\theta }{3}=-1$ are given by:
$\begin{align}
& \frac{2\theta }{3}=\frac{3\pi }{2}+2n\pi \\
& \theta =\frac{9\pi }{4}+3n\pi
\end{align}$
Where, n is any integer. All values of $\theta $ exceed $2\pi $ or are less than zero. Hence, in the interval $[0,2\pi )$ there is no solution.
