How do you solve #sin 2x + sin x = 0# over the interval 0 to 2pi?

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How do you solve #sin 2x + sin x = 0# over the interval 0 to 2pi?

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Answer 1 · 2016-04-07T05:07:51

#0, pi, (2pi)/3, (4pi)/3, 2pi#

Explanation:

sin 2x + sin x = 0
Apply the identity: 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 and x = 2pi#
b/ 2cos x + 1 = 0
#cos x = -1/2# --># x = +- (2pi)/3#
Answers for #(0, 2pi)#:
#0, pi, (2pi)/3, (4pi)/3, 2pi#
Reminder arc #(4pi)/3# is co-terminal to arc #-(2pi)/3#

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How do you solve #sin 2x + sin x = 0# over the interval 0 to 2pi?

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