How do you solve for the exact solutions in the interval [0,2pi] of #sin 2x - sin x= 0#?

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How do you solve for the exact solutions in the interval [0,2pi] of #sin 2x - sin x= 0#?

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Answer 1 · 2016-04-16T03:33:56

#0, pi/3, pi, (5pi)/3, 2pi#

Explanation:

Apply the trig identity: sin 2a = 2sin a.cos a. WE get:
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 --> #cos x = 1/2# ---> #x = = +- pi/3#
The co-terminal arc of #(-pi/3)# is #(5pi)/3#
Answers for #(0, 2pi)#:
#0, pi/3, pi, (5pi)/3, 2pi#

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How do you solve for the exact solutions in the interval [0,2pi] of #sin 2x - sin x= 0#?

1 Answer

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