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

1 Answer
Jul 22, 2016

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

Explanation:

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