Given the initial expression
[tex]\sin ^{-1}(-\frac{\sqrt[]{3}}{2})[/tex]
Set
[tex]\begin{gathered} \sin ^{-1}(-\frac{\sqrt[]{3}}{2})=x \\ \Rightarrow\sin x=-\frac{\sqrt[]{3}}{2} \end{gathered}[/tex]
On the plane, using the unitary circle
Furthermore,
[tex]\begin{gathered} \sin x=\frac{O}{H} \\ \Rightarrow\frac{O}{H}=-\frac{\sqrt[]{3}}{2}=\frac{-\frac{\sqrt[]{3}}{2}}{1} \end{gathered}[/tex]
This is a well-known triangle,
Therefore, the answer is
[tex]\Rightarrow x=-\frac{\pi}{3}[/tex]
x=-pi/3 is the answer (notice that it corresponds to the blue angle in the first figure)
Remember that the domain of arcsin(x) is [-pi/2,pi/2]
