Answer
The solutions in the interval $[0,2\pi )$ are $\frac{4\pi }{9}$, $\frac{8\pi }{9}$ and $\frac{16\pi }{9}$.
Work Step by Step
We know that the period of the secant function is $2\pi $. In the interval $\left[ 0,\left. 2\pi \right) \right.$, there are two values at which the secant function is $-2$. One is $\frac{2\pi }{3}$ and the secant is negative in quadrant II, thus the other value is:
$\begin{align}
& 2\pi -\frac{2\pi }{3}=\frac{6\pi -2\pi }{3} \\
& =\frac{4\pi }{3}
\end{align}$
Therefore, all the solutions to $\sec \frac{3\theta }{2}=-2$ are given by:
$\begin{align}
& \frac{3\theta }{2}=\frac{2\pi }{3}+2n\pi \\
& \theta =\frac{4\pi }{9}+\frac{4n\pi }{3}
\end{align}$
Or,
$\begin{align}
& \frac{3\theta }{2}=\frac{4\pi }{3}+2n\pi \\
& \theta =\frac{8\pi }{9}+\frac{4n\pi }{3}
\end{align}$
Where, n is any integer. And the solutions in the interval $[0,2\pi )$ are obtained by letting $n=0$ and $n=1$. Thus, the equation is calculated by taking first $n$ as 0 and then as 1. It can be further simplified as follows.
$\begin{align}
& \theta =\frac{4\pi }{9}+\frac{4n\pi }{3} \\
& =\frac{4\pi }{9}+\frac{4\times 0\times \pi }{3} \\
& =\frac{4\pi }{9}+0 \\
& =\frac{4\pi }{9}
\end{align}$
$\begin{align}
& \theta =\frac{4\pi }{9}+\frac{4n\pi }{3} \\
& =\frac{4\pi }{9}+\frac{4\times 1\times \pi }{3} \\
& =\frac{4\pi }{9}+\frac{4\pi }{3} \\
& =\frac{4\pi +12\pi }{9}
\end{align}$
$=\frac{16\pi }{9}$
The second equation is also computed by taking first $n$ as 0 and then as 1. It can be further simplified as follows.
$\begin{align}
& \theta =\frac{8\pi }{9}+\frac{4n\pi }{3} \\
& =\frac{8\pi }{9}+\frac{4\times 0\times \pi }{3} \\
& =\frac{8\pi }{9}+0 \\
& =\frac{8\pi }{9}
\end{align}$
$\begin{align}
& \theta =\frac{8\pi }{9}+\frac{4n\pi }{3} \\
& =\frac{8\pi }{9}+\frac{4\times 1\times \pi }{3} \\
& =\frac{8\pi }{9}+\frac{4\pi }{3} \\
& =\frac{8\pi +12\pi }{9}
\end{align}$
$=\frac{20\pi }{9}$
Since $\frac{20\pi }{9}$ is not in $[0,2\pi )$, the actual solutions in the interval $[0,2\pi )$ will be $\frac{4\pi }{9}$, $\frac{8\pi }{9}$, and $\frac{16\pi }{9}$.
