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Math Help - Find the value of x.

  1. #1
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    Find the value of x.

    For \pi<x<\frac{3\pi}{2} with cos(x)=-sin(\frac{\pi}{6}), find the value of x (do not evaluate sin(\frac{\pi}{6}) ).

    Please show all steps to any working out.

    Any help will be appreciated!
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  2. #2
    Super Member TheChaz's Avatar
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    Quote Originally Posted by Joker37 View Post
    For \pi<x<\frac{3\pi}{2} with cos(x)=-sin(\frac{\pi}{6}), find the value of x (do not evaluate sin(\frac{\pi}{6}) ).

    Please show all steps to any working out.

    Any help will be appreciated!
    Notice that -sin(\frac{\pi}{6}) = sin(\frac{-\pi}{6})
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  3. #3
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    http://latex.codecogs.com/png.latex?%5Cpi%3Cx%3C%5Cfrac{3%5Cpi}{2}

    Quote Originally Posted by Joker37 View Post
    For \pi<x<\frac{3\pi}{2} with cos(x)=-sin(\frac{\pi}{6}), find the value of x (do not evaluate sin(\frac{\pi}{6}) ).

    Please show all steps to any working out.

    Any help will be appreciated!
    sin(\pi /6)=\frac{1}{2}, so -sin(\pi /6)=-\frac{1}{2}, so we have:
    cos(x)=- \frac{1}{2}
    x= \pm acos(-\frac{1}{2})+\pi n,n\in Z
    x= \pm \frac{2\pi }{3} +\pi n,n\in Z

    if \pi<x<\frac{3\pi}{2} that x=\frac{4\pi}{3}, or x=-\frac{3\pi}{2}
    Last edited by ihavenonick; April 30th 2011 at 09:21 AM.
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  4. #4
    Super Member TheChaz's Avatar
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    Quote Originally Posted by ihavenonick View Post
    sin(π/6)=\frac{1}{2}, so -sin(п/6)=\frac-{1}{2}, so we have:
    cos(x)=\frac-{1}{2}
    x=\pm acos(-1/2)+πn,n\in Z
    x=\pm 2π/3 +πn,n\in Z

    if \pi<x<\frac{3\pi}{2} that x=\frac-{3\pi}{2}, or x=\frac-{3\pi}{2}
    Besides doing the entire problem, you evaluated sin(π/6) - something that was expressly forbidden!
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  5. #5
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    Quote Originally Posted by TheChaz View Post
    Notice that -sin(\frac{\pi}{6}) = sin(\frac{-\pi}{6})
    sure thing!
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  6. #6
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    Quote Originally Posted by TheChaz View Post
    Besides doing the entire problem, you evaluated sin(π/6) - something that was expressly forbidden!
    I was editing the post at that moment. Look at this now.
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  7. #7
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    Quote Originally Posted by ihavenonick View Post
    sin(\pi /6)=\frac{1}{2}, so -sin(\pi /6)=-\frac{1}{2}, so we have:
    cos(x)=- \frac{1}{2}
    x= \pm acos(-\frac{1}{2})+\pi n,n\in Z
    x= \pm \frac{2\pi }{3} +\pi n,n\in Z

    if \pi<x<\frac{3\pi}{2} that x=\frac{4\pi}{3}, or x=-\frac{3\pi}{2}
    Actually the answer is meant to be \frac{7\pi}{6} but nevermind I think I understand how to do it now.
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  8. #8
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    Quote Originally Posted by ihavenonick View Post
    sin(\pi /6)=\frac{1}{2}, so -sin(\pi /6)=-\frac{1}{2}, so we have:
    cos(x)=- \frac{1}{2}
    x= \pm acos(-\frac{1}{2})+\pi n,n\in Z
    x= \pm \frac{2\pi }{3} +\pi n,n\in Z

    if \pi<x<\frac{3\pi}{2} that x=\frac{4\pi}{3}, or x=-\frac{3\pi}{2}
    I`m sorry I`ve done a mistake. Here is a correct answer:
    sin(\pi /6)=\frac{1}{2}, so -sin(\pi /6)=-\frac{1}{2}, so we have:
    cos(x)=- \frac{1}{2}
    x= \pm acos(-\frac{1}{2})+\pi n,n\in Z
    x= \pm \frac{2\pi }{3} +\pi n,n\in Z

    if \pi<x<\frac{3\pi}{2} that x=\frac{4\pi}{3}(or x=-\frac{2\pi}{3})
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  9. #9
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    Jet another:
    cos(x)=- -sin(\pi /6)
    x= \pm acos(-sin(\pi /6))+\pi n,n\in Z; cos(3\pi /2 - \alpha )=-\sin \alpha
    x= \pm acos(cos(3\pi /2 - \pi /6)+\pi n,n\in Z
    x= \pm acos(cos(9\pi /6-\pi /6)+\pi n,n\in Z
    x= \pm acos(cos(8\pi /6)+\pi n,n\in Z
    x= \pm acos(cos(4\pi /3)+\pi n,n\in Z
    x= \pm 4\pi /3 +\pi n,n\in Z

    if \pi<x<\frac{3\pi}{2} that x=\frac{4\pi}{3}

    P.S. I don`t understand too. If calculate, the correct answer is \frac{7\pi}{6} as you wrote.
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