solve for between 0 and 2 pal
cos^2 (2X)=cos (2X)
thanks in advance
[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
Yes, but you could do "cos 2x= 0" separately. If cos 2x is NOT 0, divide by it to get cos 2x= 1 leading to $\displaystyle 2x= 2n\pi$and so $\displaystyle x= n\pi$, for any integer n. If cos 2x is equal to 0 then $\displaystyle 2x= \frac{\pi}{2}+ n\pi$ so $\displaystyle x= \frac{\pi}{4}+ n\frac{\pi}{2}$.