Question

How do you find the exact solutions of the equation `sin2x-sinx=0` in the interval `[0,2pi)`?

Answer 1

`0, pi/3, pi, (5pi)/3`,

Explanation:

Use trig identity: sin 2 x = 2sin x.cos x.
Replace in the equation sin 2x = 2sin x.cos x
2sin x.cos x - sin x = 0
sin x(2cos x - 1) = 0
a. sin x = 0 --> `x = 0, x = pi, x = 2pi`
b. 2cos x - 1 = 0 --> `cos x = 1/2`
Unit circle gives:
`x = 1/2` --> `cos x = +- pi/3`
Note. `- pi/3` and `(5pi)/3` are co-terminal.
Answers for `(0, 2pi)`
`0, pi/3, pi, (5pi)/3, 2pi`