Solve equation: (Is my answer correct ?)

#sin^2x + sin^2 2x = sin^2 3x#
#sin^2x + (2sinxcosx)^2 = [sinx(3-4sin^2x)] #
#sin^2x +4sin^2xcos^2x = sin^2x(9-24sin^2x +16sin^4x) #
#sin^2x +4sin^2xcos^2x = 16sin^6x - 24sin^4x +9sin^2x#
#4sin^2x(4sin^4x - 6sin^2x +2 - cos^2x) =0 #
#sin^2x = 0#
#x=kπ#
⋁ #4sin^4x - 6sin^2x+2-1+sin^2x=0 #
#4sin^4x - 5sin^2x +1 =0#
#sin^2x = t#
#4t^2-5t+1=0 #
∆=9
#t_1 = 1/4# and #t_2 =1#

so...

for #t_1 = 1/4# :
#1/2 = sinx# or #-1/2 = sinx #
#x=π/6 + 2kπ or x=-π/6 + 2kπ#

for #t_2 = 1# :
#sinx =1 or sinx =-1#
#x=π/2+2kπ or x=-π/2+2kπ #

# x=π/6 + kπ ⋁ x=π/2 + kπ#

1 Answer
Nghi N.
Jun 27, 2018

#x = +- 37^@76 + k180^@#
#x = kpi#, and #x = pi/2 + kpi#

Explanation:

#sin^2 x - sin^2 3x + sin^2 2x = 0#
#(1 - cos 2x) - (1 - cos 6x) + sin^2 2x = 0#
#(cos 6x - cos 2x) + sin^2 2x = 0#
#- 2sin 4x.sin 2x + sin^2 2x = 0#
#sin 2x (sin 2x - 2sin 4x) = sin 2x(sin 2x - 4sin2x.cos 2x) = 0 #
#sin^2 2x(1 - 4cos 2x) = 0#
Either factor should be zero
.
a. #sin 2x = 0#. Unit circle gives:
1. #2x = 2kpi# --> #x = kpi#
2. #2x = pi + 2kpi# --> #x = pi/2 + kpi#
b. #4cos 2x = 1# --> #cos 2x = 1/4#
Calculator and unit circle give -->
1. #2x = +- 75^@52 + k360^@# -->
#x = +-37^@76 + k180^@#

Related questions
Impact of this question
996 views around the world
You can reuse this answer
Creative Commons License