Answer
$$s = \frac{{11\pi }}{6}$$
Work Step by Step
$$\eqalign{
& \left[ {\frac{{3\pi }}{2},2\pi } \right];{\text{ }}\sin s = - \frac{1}{2} \cr
& {\text{From the unit circle we can see that in the interval }}\left[ {\frac{{3\pi }}{2},2\pi } \right],{\text{ }} \cr
& {\text{the arc length }}s = \frac{{11\pi }}{6}{\text{ is associate with the point }}\left( {\frac{{\sqrt 3 }}{2}, - \frac{1}{2}} \right). \cr
& {\text{The second coordinate is }} \cr
& \sin s = \sin \frac{{11\pi }}{6} = - \frac{1}{2} \cr
& {\text{then}}\,\,s = \frac{{11\pi }}{6} \cr} $$
