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Math Help - Algebra

  1. #1
    Junior Member rednest's Avatar
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    Algebra

    Solve the equation:

    cos x+\sqrt{1-\tfrac{1}{2} sin2x} = 0, in the interval 0^\circ \le x < 360^\circ

    Please help me on how to simplify this equation in the exam (any techniques?).

    Thanks!
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by rednest View Post
    Solve the equation:

    cos x+\sqrt{1-\tfrac{1}{2} sin2x} = 0, in the interval 0^\circ \le x < 360^\circ

    Please help me on how to simplify this equation in the exam (any techniques?).

    Thanks!
    i suggest you may start with..
    \sqrt{1-\tfrac{1}{2} \sin 2x} = -\cos x
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  3. #3
    Super Member Aryth's Avatar
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    Here's what I did. I'm pretty sure it's right:

    We have-

    \cos{x} + \sqrt{1 - \frac{1}{2}\sin{2x}} = 0

    First:

    \sqrt{1 - \frac{1}{2}\sin{2x}} = -\cos{x}

    Next:

    1 - \frac{1}{2}\sin{2x} = \cos^2{x}

    Keep in mind that (-\cos{x})^2 = \cos^2{x}

    Then:

    -\frac{1}{2}\sin{2x} = \cos^2{x} - 1

    Now we remember that:

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

    Therefore:

    \cos^2{x} - 1 = -\sin^2{x}

    So:

    -\frac{1}{2}\sin{2x} = -\sin^2{x}

    Negatives can cancel, giving us:

    \frac{1}{2}\sin{2x} = \sin^2{x}

    Now, we go to another identity:

    \sin{2x} = 2\sin{x}\cos{x}

    So:

    \sin{x}\cos{x} = \sin^2{x}

    \left(\frac{1}{2}(2\sin{x}\cos{x}) = \sin{x}\cos{x}\right)

    Now:

    \sin{x} = \sin{x}\tan{x}

    And finally:

    1 = \tan{x}

    x = 45^{\circ}, \ 225^{\circ}

    And there you go.
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Aryth View Post

    Now:

    \sin{x} = \sin{x}\tan{x}

    And finally:

    1 = \tan{x}

    x = 45^{\circ}, \ 225^{\circ}

    And there you go.
    we cannot do this immediately, since x can be zero (and pi) and so does \sin x..

    so, the best way is to do this: \sin{x}\tan{x} - \sin x = (\sin x) (\tan x - 1) = 0..

    and there you go.. 1 = \tan{x} and also \sin x = 0

    after finding the values of x, try to check whether it fits the original equation..
    Last edited by kalagota; June 23rd 2008 at 01:05 AM.
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