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Math Help - help...

  1. #1
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    help...

    ok so i got this problem and i did a bunch of things to try and solve it and i just cant seem to get it....

    Cos(X)*Cos(2x)=1

    your help would be appreciated. thanks.
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  2. #2
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    Quote Originally Posted by mathg33k View Post
    ok so i got this problem and i did a bunch of things to try and solve it and i just cant seem to get it....

    Cos(X)*Cos(2x)=1

    your help would be appreciated. thanks.
    well, it's not very easy to be done by hand ...

    use the identity \cos(2x) = 2\cos^2{x} - 1

    \cos(x)(2\cos^2{x} - 1) = 1

    2\cos^3{x} - \cos{x} - 1 = 0

    one obvious solution (and there is no other way to find it other than "observation") is cos{x} = 1

    so, two solutions for x are x = 0 or x = 2\pi.

    investigating other possible solutions ...

    use synthetic division with the coefficients ...

    Code:
    1].........2..........0.........-1.........-1
    .....................2...........2..........1
    -----------------------------------------------
    ...........2..........2..........1..........0
    the depressed polynomial is ...

    2\cos^2{x} + 2\cos{x} + 1 = 0

    since b^2-4ac < 0, there are no other real solutions for \cos{x}

    ... told you it wasn't "nice".
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  3. #3
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    ok thanks that helps a lot.
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  4. #4
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    notice

    if cos (x)=1 this means x=0,360,2*360,.....
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  5. #5
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    Quote Originally Posted by ekledes View Post
    notice

    if cos (x)=1 this means x=0,360,2*360,.....
    note that the given solutions were in the interval 0 to 2\pi radians.

    if you're trying to list possible solutions in degrees outside of that primary interval, then list them all ...

    x \in \{... \, , \, -2\cdot 360, -360, 0 , 360, 2 \cdot 360, \, ... \, \}
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