How do I simplify arcsin (sin 6 pi) given the interval 0 ≤ theta < 2pi

Answer 1

since sin(?) and arcsin(? are inverses of each other,

arcsin( sin (6π) ) = 6π , but that is beyond your stated boundary.

so arcsin(sin 6π)
= arcsin 0
= 0 , π , 2π