How do you solve #-3cosx+3=2sin^2x# in the interval #0<=x<=2pi#?

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Answer 1 · 2016-11-07T09:56:40

#x={(2pi)/3,pi,(4pi)/3}#

#3cosx+3=2sin^2x#

#hArr3cosx+3=2(1-cos^2x)#

or #3cosx+3=2-2cos^2x#

or #2cos^2x+3cosx+1-0#

or #2cos^2x+2cosx+cosx+1-0#

or #2cosx(cosx+1)+1(cosx+1)=0#

or #(2cosx+1)(cosx+1)=0#

and hence either #2cosx+1=0# i.e. #cosx=-1/2#

or #cosx+1=0# i.e. #cosx=-1#

Considering #0<= x <= 2pi#,

#x=(2pi)/3# or #x=(4pi)/3# or #x=pi#

i.e. #x={(2pi)/3,pi,(4pi)/3}#