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Thread: Algebra

  1. #1
    Junior Member rednest's Avatar
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    Algebra

    Solve the equation:

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

    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:

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

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

    Thanks!
    i suggest you may start with..
    $\displaystyle \sqrt{1-\tfrac{1}{2} \sin 2x} = -\cos x$
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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-

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

    First:

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

    Next:

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

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

    Then:

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

    Now we remember that:

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

    Therefore:

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

    So:

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

    Negatives can cancel, giving us:

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

    Now, we go to another identity:

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

    So:

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

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

    Now:

    $\displaystyle \sin{x} = \sin{x}\tan{x}$

    And finally:

    $\displaystyle 1 = \tan{x}$

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

    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:

    $\displaystyle \sin{x} = \sin{x}\tan{x}$

    And finally:

    $\displaystyle 1 = \tan{x}$

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

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

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

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

    after finding the values of $\displaystyle x$, try to check whether it fits the original equation..
    Last edited by kalagota; Jun 23rd 2008 at 12:05 AM.
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