How do you solve #4cos^2 (4x) -3=0#?

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Answer 1 · 2016-07-22T09:43:07

#x=(npi)/2+-pi/24# or #x=(npi)/2+-(5pi)/24#

#4cos^2(4x)-3=0#

#hArrcos^2(4x)-3/4=0#

#hArrcos^2(4x)-(sqrt3/2)^2=0#

#hArr(cos(4x)-(sqrt3/2))(cos(4x)+(sqrt3/2))=0#

i.e. either #cos(4x)-(sqrt3/2)=0# i.e. #cos(4x)=sqrt3/2=cos(pi/6)#

and #4x=2npi+-pi/6# i.e. #x=(npi)/2+-pi/24#

or #cos(4x)+(sqrt3/2)=0# i.e. #cos(4x)=(-sqrt3/2)=cos((5pi)/6)#

and #4x=2npi+-(5pi)/6# i.e. #x=(npi)/2+-(5pi)/24#