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Math Help - Trig equations

  1. #1
    Member classicstrings's Avatar
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    Trig equations

    How do I solve this?

    -0.5 = sin((Pi*x)/12)

    Sin is -0.5 when the angle is 30. Since it is negative, angles are 210 and 330.

    I get eventually 7,11,14,23...

    Which are wrong somehow. Please help! Thx
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  2. #2
    Eater of Worlds
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    Solve for x:

    sin^{-1}(-1/2)=\frac{{\pi}x}{12}

    x=\frac{12sin^{-1}(-1/2)}{\pi}=-2

    \frac{{\pi}(-2)}{12}=\frac{-\pi}{6}

    sin(\frac{-5\pi}{6}+2n{\pi})=-1/2

    sin(\frac{-\pi}{6}+2n{\pi})=-1/2, n=0,1,2,......
    Last edited by galactus; July 20th 2006 at 11:29 AM. Reason: typo
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  3. #3
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    Quote Originally Posted by classicstrings
    How do I solve this?

    -0.5 = sin((Pi*x)/12)

    Sin is -0.5 when the angle is 30. Since it is negative, angles are 210 and 330.

    I get eventually 7,11,14,23...

    Which are wrong somehow. Please help! Thx
    If you simply had,
    \sin x =-\frac{1}{2}
    Then, then the two coterminal angles are,
    x=\frac{4\pi}{3}
    x=\frac{5\pi}{3}
    Then all solutions are,
    x=\frac{4\pi}{3}+2\pi k
    x=\frac{5\pi}{3}+2\pi k
    But since you have,
    \sin \left( \frac{\pi x}{12} \right) =-\frac{1}{2}
    All solutions are,
    \frac{\pi x}{12}=\frac{4\pi}{3}+2\pi k
    \frac{\pi x}{12}=\frac{5\pi}{3}+2\pi k
    Thus, (mutiply by reciprocal)
    x=16+24k
    x=20+24k
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    x=16+24k
    x=20+24k
    k=0,\ \pm 1,\ \pm 2, ..
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  5. #5
    Member classicstrings's Avatar
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    Thanks a lot guys! I realised today when I logged on that you changed the updated the number of (1800+) members sign too!
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