How to solve this?#sinx+sin2x+sin3x=0#

1 Answer
Nghi N.
Mar 22, 2017

#x = (kpi)/2#
#x = +- (2pi)/3 + 2kpi#

Explanation:

Use trig identity:
#sin a + sin b = 2sin ((a + b)/2).cos ((a - b)/2)#
In this case:
#sin x + sin 3x = 2sin (2x).cos (x)#
#(sin x + sin 3x) + sin 2x = 2sin (2x)cos (x) + sin (2x) = #
#= sin (2x)(2cos x + 1) = 0#
Either one of the 2 factors must be zero.
a. sin 2x = 0
2x = 0 --> x = 0
#2x = pi# --> #x = pi/2#
#2x = 2pi# --> #x = pi#
General answer: #x = (kpi)/2#
b. 2cos x + 1 = 0 --> #cos x = - 1/2#
Trig table and unit circle give 2 solutions -->
#x = +- (2pi)/3 + 2kpi#

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