How do you solve sin 2x + sin x = 0sin2x+sinx=0 over the interval 0 to 2pi?

1 Answer
Nghi N.
Apr 7, 2016

0, pi, (2pi)/3, (4pi)/3, 2pi0,π,2π3,4π3,2π

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 = 2pix=0,x=πandx=2π
b/ 2cos x + 1 = 0
cos x = -1/2cosx=−12 --> x = +- (2pi)/3x=±2π3
Answers for (0, 2pi)(0,2π):
0, pi, (2pi)/3, (4pi)/3, 2pi0,π,2π3,4π3,2π
Reminder arc (4pi)/34π3 is co-terminal to arc -(2pi)/3−2π3

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