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Math Help - Trigonometric function

  1. #1
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    Trigonometric function

    Hello guys!

    I have a problem solving this equation. I really dont know how to start Could someone pls help me with this...

    The problem is that Find all solutions of the given equation in the indicated interval.

    3sin2X+cosX = 0 [-pi/2, pi/2]

    Thanks in advanced.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi,

    Hint : \sin (2x)=2\sin x\cos x.
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  3. #3
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    thank you very much for replying.

    OK so using that identity my equation

    from 3sin2X+cox

    will become (3)(2)sinxcosx + cosx = 0

    6sinxcosx + cox = 0

    Im I doing it right?
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  4. #4
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by cjru View Post
    thank you very much for replying.

    OK so using that identity my equation

    from 3sin2X+cox

    will become (3)(2)sinxcosx + cosx = 0

    6sinxcosx + cox = 0

    Im I doing it right?
    Yes then factor out a cos x
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  5. #5
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    Quote Originally Posted by 11rdc11 View Post
    Yes then factor out a cos x
    ok then I get cosX (6sinX) = 0

    Therefore Cos X = 0 where in Cos can only be zero at x equal to pi/2 and -pi/2

    Is this right?

    How about the 6sinx = 0

    Pls check if i did it right....

    sin x = 0/6 then sin x = 0

    Therefore sin x = 0 wherein sin can only be zero at x equal to 0 and pi.
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  6. #6
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    not quite ...

    \cos{x}(6\sin{x} + 1) = 0
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  7. #7
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    o yeah! I over look it. So what will be the solution for 6sinX+1?

    If I equate that to zero sinx = 1/6 Is this correct?
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  8. #8
    Super Member 11rdc11's Avatar
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    ~

    Quote Originally Posted by cjru View Post
    o yeah! I over look it. So what will be the solution for 6sinX+1?

    If I equate that to zero sinx = 1/6 Is this correct?
    \sin{x} = -~\frac{1}{6}

    x = \sin^{-1} {\bigg(-~\frac{1}{6}\bigg)}
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  9. #9
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    Thank you so much. Math help forum is the best.

    God bless and more power to you guys!

    Again thank you
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