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Answer 1 · 2018-02-16T07:07:01

The answers are #x=pi/3,pi,(5pi)/3#.

#sin^2x=cos^2(x/2)#

#sin^2x=(cos(x/2))^2#

#sin^2x=(sqrt((1+cosx)/2))^2#

#sin^2x=(1+cosx)/2#

#2sin^2x=1+cosx#

#2(1-cos^2x)-cosx-1=0#

#2-2cos^2x-cosx-1=0#

#-2cosx^2x-cosx+1=0#

Let #u = cosx#:

#-2u^2-u+1=0#

#2u^2+u-1=0#

#(2u-1)(u+1)=0#

#u=-1,1/2#

Substitute #cosx# back in for #u#:

#cosx=-1,cosx=1/2#

The solution for #cosx=-1# is #pi#, and solutions for #cosx=1/2# are #pi/3# and #(5pi)/3#. The final solution set is:

#x=pi/3,pi,(5pi)/3#