How to solve #4sin^2((3x+pi)/6)=3#?

So, #sin((3x+pi)/6)=+-sqrt3/2#. Then what?

2 Answers
Roy Ø.
Jun 7, 2018

#x=(2k+1)pi#v#(2k-1/3)pi#

Explanation:

We have #sin(2pi/3)=sqrt3/2# (see https://socratic.org/questions/what-is-sin-2pi-3-equal-to)

Therefore
#sin((3x+pi)/6)=+-sin(2pi/3)#
#(3x+pi)/6=(2/3pi +2kpi)# v #(-2/3pi +2kpi)#
#-2/3pi# should have the same sin value as #4/3pi#
so that we get
#3x+pi=6(2/3pi +2kpi)# v #6(4/3pi +2kpi)#
#3x=(4pi-pi+12kpi)#v#(12pi-pi+12kpi)#
#3x=(3pi+12kpi)#v#(11pi+12kpi)#
#x=(pi+4kpi)# v #(11/3pi+4kpi)#
#x=(2k+1)pi# v #(2k-1/3)pi#

Nghi N.
Jun 8, 2018

#x = (2k + 1)pi/3#

Explanation:

#4sin^2 ((3x + pi)/6) = 3#
#sin ((3x + pi)/6) = +- sqrt3/2#
a. #sin ((3x + pi)/6) = sqrt3/2#
Trig table and unit circle give 2 solutions for #(3x + pi)/6#:
1. #(3x + pi)/6 = pi/3# --> #3x + pi = 2pi#
#3x = 2pi - pi = pi + 2kpi# -->#x = pi/3 + (2kpi)/3#
2. #(3x + pi)/6 = (2pi)/3# --> #3x + pi = 4pi#
#3x = 3pi + 2kpi# --> #x = pi + (2kpi)/3#
General answer: #x = (2k + 1)pi/3#
b. #sin ((3x + pi)/6) = - sqrt3/2#
Trig table and unit circle give 2 solutions for #(3x + pi)/6#
1. #(3x + pi)/ 6 = (4pi/3)# --> #(3x + pi) = 8pi#
#3x = 7pi + 2kpi# --> #x = (7pi)/3 + (2kpi)/3#, or
#x = pi/3 + (2kpi)/3#
2. #(3x + pi)/6 = (5pi)/3# --> #3x + pi = 10pi#
#3x = 9pi + 2kpi# --> #x = 3pi + (2kpi)/3#, or
x = pi + (2kpi)/3
General answer: #x = (2k + 1)pi/3#

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