# How do you find the exact value of sin^-1 (sqrt3/2)?

pi/3 = 60º
$\frac{\sqrt{3}}{2} = \sin x$
$x = \frac{\pi}{3} + 2 k \pi$ or $x = \pi - \frac{\pi}{3} + 2 k \pi$ , for all $k \setminus \in m a t h \boldsymbol{Z}$
But the function $y = \arcsin x$ returns a single angle in the interval $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$.