solve for between 0 and 2 pal

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

thanks in advance

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- Jan 12th 2010, 06:11 PMalessandromangionesolve given general solution
solve for between 0 and 2 pal

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

thanks in advance - Jan 12th 2010, 06:14 PMpickslides
$\displaystyle \cos^2 (2X)=\cos (2X)$

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

$\displaystyle \cos (2X)=1$

Can you solve it from here? - Jan 12th 2010, 07:03 PMalessandromangione
thank u very much...now i can do it!

- Jan 12th 2010, 07:04 PMdifferentiate
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, 07:34 PMnikhil
[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 is0 to 2pi**between**

therefor finally

x= pi/4,3pi/4,pi - Jan 13th 2010, 02:35 AMHallsofIvy
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 $\displaystyle x= \frac{\pi}{4}+ n\frac{\pi}{2}$.