Question

How do you find the exact value of `sin2theta-costheta=0` in the interval `0<=theta<360`?

Answer 1

`pi/6; pi/2; (3pi)/2; (5pi)/6`

Explanation:

Replace sin 2t by 2sint.cos t -->
2sin t.cos t - cos t = 0
cos t(2sin t - 1) = 0
Use trig table of special arcs and unit circle -->
a. cos t = 0 --> arc `t = pi/2, and t = (3pi)/2`
b. (2sin t - 1) = 0 --> sin t = 1/2 -->
`t = pi/6 and t = (5pi)/6`
Answers for `(0, 2pi)`:
`pi/6, pi/2, (3pi)/2, (5pi)/6`