Answer
$\sin ^{-1}\left( \dfrac {\sqrt {3}}{2}\right) =$$\dfrac {\pi }{3}+2\pi k;=\dfrac {2\pi }{3}+2\pi k$
Work Step by Step
Function $sine$ has a period of $2\pi$
It means we must find all the solutions in $[0,2\pi) $ and then add $2\pi k$ to it (k is integer number)
$\sin ^{-1}\left( \dfrac {\sqrt {3}}{2}\right) =$$\dfrac {\pi }{3}+2\pi k;=\dfrac {2\pi }{3}+2\pi k$
