Question

How do you solve the equation `sin^2theta-sintheta=0` for `[0,2pi)`?

Answer 1

`0, pi/2, pi, 2pi`

Explanation:

`sin^2 t - sin t = 0`
`sin t(sin t - 1) = 0`
Use unit circle -->
a. sin t = 0 --> 3 solution arcs -->
arc t = 0, arc `t = pi`, and arc `t = 2pi`

b. sin t - 1 = 0 --> sin t = 1 --> arc `t = pi/2`

Answers for`(0, 2pi)`
`0, pi/2, pi, 2pi`