# Thread: solve given general solution

1. ## solve given general solution

solve for between 0 and 2 pal

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

2. $\displaystyle \cos^2 (2X)=\cos (2X)$

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

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

Can you solve it from here?

3. thank u very much...now i can do it!

4. 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...

5. [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

6. 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 $\displaystyle 2x= 2n\pi$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 $\displaystyle x= \frac{\pi}{4}+ n\frac{\pi}{2}$.