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Math Help - More Trigonometric Help Need!

  1. #1
    nee
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    More Trigonometric Help Need!

    solve the equation for solutions over the interval [0, 360)

    4cos(squared)x + 4cosx = 1

    I'm completely lost on this one!
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  2. #2
    Moo
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    Hi !
    Quote Originally Posted by nee View Post
    solve the equation for solutions over the interval [0, 360)

    4cos(squared)x + 4cosx = 1

    I'm completely lost on this one!
    For you to better understand, substitute u=\cos(x)

    The equation is now 4u^2+4u=1, that is to say 4u^2+4u-1=0

    Now, use the method of discriminant or complete the square to get the solution

    A slightly different way to do it is to substitute u=2 \cos(x) ^^
    Last edited by Moo; August 25th 2008 at 12:32 PM.
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  3. #3
    nee
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    You mean like this?

    4cos(squared)x + 4cosx = 1
    4cos(squared)x + 4cosx + 1 = 1 + 1
    4cos(squared)x + 4cosx +1 = 2
    (2cosx + 1)(2cosx + 1) = 2

    but then what?

    would the following be correct?:

    2cosx = 2 or 2cosx = 2
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by nee View Post
    You mean like this?

    4cos(squared)x + 4cosx = 1
    4cos(squared)x + 4cosx + 1 = 1 + 1
    4cos(squared)x + 4cosx +1 = 2
    (2cosx + 1)(2cosx + 1) = 2

    but then what?

    would the following be correct?:

    2cosx = 2 or 2cosx = 2
    no. as Moo said, the equation you are using is 4 \cos^2 x + 4 \cos x - 1 = 0

    that is what you should be solving. she replaced cos(x) with you to make things look easier. just work on solving the quadratic equation she gave you. you can use the quadratic formula
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  5. #5
    Moo
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    Quote Originally Posted by nee View Post
    You mean like this?

    4cos(squared)x + 4cosx = 1
    4cos(squared)x + 4cosx + 1 = 1 + 1
    4cos(squared)x + 4cosx +1 = 2
    (2cosx + 1)(2cosx + 1) = 2

    but then what?

    would the following be correct?:

    2cosx = 2 or 2cosx = 2
    Eh !!!
    That's good ! It's just like completing the square

    (2cos(x)+1)(2cos(x)+1)=(2cos(x)+1)=2
    Thus 2 \cos(x)+1=\pm \sqrt{2}
    And go from here
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  6. #6
    nee
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    Ok howbout this then:

    cosx = -4 +or- (squareroot)4(squared) - 4(4)(-1) / 2(4)

    = -4 +or- (squareroot)0 / 8

    = -4 / 8

    = -1/2

    therefore solution set is:

    {120, 240}

    ???
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