Solve equation: (Is my answer correct ?)

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1 Answer
Jun 27, 2018

#x = kpi; x = pi/2 + kpi#
#x = pi/6 + (kpi)/3#

Explanation:

#sin^2 2x = sin^2 3x - sin^2 x#(1)
Develop the right side:
RS = (sin 3x - sin x)(sin 3x + sin x)
RS = (2cos 2x.sin x)(2sin 2x.cos x) = (2sin x.cos x)(2sin 2x.cos 2x)=
RS = sin 2x.sin 4x
Equation (1) becomes:
#sin^2 2x = sin 2x.sin 4x sin^2 2x - sin 2x sin 4x = 0#
#sin 2x(sin 2x - sin 4x) = 0#
Either factor should be zero.

a. sin 2x = 0. Unit circle gives:
#2x = 2kpi #--> #x = kpi#
#2x = pi + 2kpi# --> #x= pi/2 + kpi#

b. # sin 4x = sin 2x #
Unit circle gives:
#4x = 2x + 2kpi#, and #4x = pi - 2x + 2kpi#
1. #4x = 2x + 2kpi# --> #2x = 2kpi# --> #x = kpi#
2. #4x = pi - 2x + 2kpi# --> #6x = pi + 2kpi#
#x = (pi)/6 + (kpi)/3#