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Math Help - solve given general solution

  1. #1
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    solve given general solution

    solve for between 0 and 2 pal

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

    thanks in advance
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  2. #2
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     \cos^2 (2X)=\cos (2X)

    divide both sides by  \cos (2X) gives

     \cos (2X)=1

    Can you solve it from here?
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  3. #3
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    thank u very much...now i can do it!
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  4. #4
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    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...
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  5. #5
    Senior Member nikhil's Avatar
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    Lightbulb

    [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
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  6. #6
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    Quote Originally Posted by differentiate View Post
    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\piand 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}.
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