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

1 Answer
Feb 20, 2017

#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#