divide both sides by gives
Can you solve it from here?
[cos(2x)]^2=cos(2x)
or
[cos(2x)]^2- cos(2x)=0
cos(2x)[cos(2x)-1]=0
cos(2x)=0 or cos(2x)=1
a) if cos(2x)=0 then
2x=pi/2,3pi/2 or
x=pi/4,3pi/4
b)cos(2x)=1 then
2x=0,2pi or
x=0,pi
hence
x=0,pi/4,3pi/4,pi
now 0 may be excluded as interval is between 0 to 2pi
therefor finally
x= pi/4,3pi/4,pi