How do you solve for the exact solutions in the interval [0,2pi] of #2 + cos^2(2x)=3 sin^2(2x)#?

Question

1741 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-23T08:41:50

#x=npi+-pi/3# or #x=npi+-pi/6#

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

or #2+cos^2(2x)=3-3cos^2(2x))#

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

or #4cos^2(2x)-1=0#

or #(2cos(2x)+1)(2cos(2x)-1)=0#

i.e. #cos(2x)=-1/2# i.e. #2x=2npi+-(2pi)/3#

i.e. #x=npi+-pi/3#

or #cos(2x)=1/2# i.e. #2x=2npi+-pi/3#

i.e. #x=npi+-pi/6#