I am trying to understand how to do this problem. Can someone guide me please?
Here is the problem:
cos^2 (2x) +2sin^2(2x)=2 Here is what I have tried: I made the equation as (cos2x)^2+2(sin2x)^2=2 Then, cos2x=1-2sin^2 x sin2x=2sinxcosx (1-2sin^2 x)^2 + 2(2sinxcosx)^2=2 (1-4sin^2 x+ 4sin^4x+8sin^2x(1-sin^2 x) 1-sin^2 x +4sin^4 x + 8sin^2 x -8sin^4 x=2 1-7sin^2x-4sin^4x=2 -4sin^4x-7sin^2x+1-2=0 -4sin^4x -7sin^2x-1=0
then, I am either stuck or I have a wrong solution.