NEED ASAP! use the unit circle definition of the trigonometric functions to solve the inequality sin θ - sqrt3/2 >0 on the interval [0,2pi)

Question

sin θ - √3 /2 >0 on interval [0,2π)

a sketch of the unit circle and the line y=√3/2 is helpful

Answer 1

sinx - sqr3/2 > 0, with 0<x<2pi

sinx > sqr3/2 (that's (sqr3)/2 not sqr(3/2))

x > 60 degrees or pi/3 and x < 120 or 2pi/3

pi/3 < x < 2pi/3 radians

or 60 < x < 120 degrees