# solve given general solution

• Jan 12th 2010, 07:11 PM
alessandromangione
solve given general solution
solve for between 0 and 2 pal

cos^2 (2X)=cos (2X)

• Jan 12th 2010, 07:14 PM
pickslides
$\cos^2 (2X)=\cos (2X)$

divide both sides by $\cos (2X)$ gives

$\cos (2X)=1$

Can you solve it from here?
• Jan 12th 2010, 08:03 PM
alessandromangione
thank u very much...now i can do it!
• Jan 12th 2010, 08:04 PM
differentiate
I thought you weren't allowed to divide by cos2x since cos2x might be equal to zero, in which if you divide by cos2x, there will be no solution...
• Jan 12th 2010, 08:34 PM
nikhil
[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
• Jan 13th 2010, 03:35 AM
HallsofIvy
Quote:

Originally Posted by differentiate
I thought you weren't allowed to divide by cos2x since cos2x might be equal to zero, in which if you divide by cos2x, there will be no solution...

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 $2x= 2n\pi$and so $x= n\pi$, for any integer n. If cos 2x is equal to 0 then $2x= \frac{\pi}{2}+ n\pi$ so $x= \frac{\pi}{4}+ n\frac{\pi}{2}$.