Answer
$\theta=-60^o$
Work Step by Step
RECALL :
(1) For $y=\csc^{-1}{(x)}\longrightarrow \csc{y}=x,$ where $y$ must be in the interval $[-90^o, 0) \cup (0, 90^o]$.
(2) $\csc{\theta} = \frac{1}{\sin{\theta}}$
(3) $y=\sin^{-1}{(x)} \longrightarrow \sin{y}=x$, where $y$ is in the interval $[-90^o, 90^o]$
Thus,
$\theta=\csc^{-1}{\left(-\frac{2\sqrt3}{3}\right)}\longrightarrow \csc{\theta}=-\frac{2\sqrt3}{3}$
Using the definition in (2) above gives:
\begin{array}{ccc}
&\csc{\theta} &= &-\frac{2\sqrt3}{3}
\\&\frac{1}{\sin{\theta}}&=&-\frac{2\sqrt3}{3}
\\&1(3) &=&-2\sqrt3\cdot (\sin{\theta})
\\&3 &=&-2\sqrt3\cdot (\sin{\theta})
\\&\frac{3}{-2\sqrt3}&=&\sin{\theta}
\\&-\frac{\sqrt3}{2}&=&\sin{\theta}
\\&\theta&=&\sin^{-1}{(-\frac{\sqrt3}{2})}
\end{array}
Use a scientific calculator's inverse tangent function in degree mode to obtain:
$\theta = \sin^{-1}{(-\frac{\sqrt3}{2})}
\\\theta=-60^o$
