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Math Help - inverse cos

  1. #1
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    inverse cos

    The question reads Let f(x) = 6 cos^-1 x + 2009pi

    Find x values such that f^-1 (x) = (sq rt 3)/2

    So I am thinking that with the cos ^-1 that the range for x is 0 to pi so that it is a one to one function.

    Then what I am thinking is to find

    6 cos^-1 x + 2009pi = (sq rt 3)/2 to find out what x leads to (sq rt 3)/2

    But where do I go from here? I don't know how to handle the cos ^-1 x.

    Thanks
    Calculus Beginner
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  2. #2
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    Quote Originally Posted by calcbeg View Post
    The question reads Let f(x) = 6 cos^-1 x + 2009pi

    Find x values such that f^-1 (x) = (sq rt 3)/2

    So I am thinking that with the cos ^-1 that the range for x is 0 to pi so that it is a one to one function.

    Then what I am thinking is to find

    6 cos^-1 x + 2009pi = (sq rt 3)/2 to find out what x leads to (sq rt 3)/2

    But where do I go from here? I don't know how to handle the cos ^-1 x.

    Thanks
    Calculus Beginner
    Use the following basic fact about one-to-one functions and their inverse:

    f(a) = b \Rightarrow f^{-1}(b) = a.
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  3. #3
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    HI

    Thanks for the formula but what do I do with the cos^-1 x? How do I figure out what x equals?

    Thanks
    Calculus Beginner
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  4. #4
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    Quote Originally Posted by calcbeg View Post
    HI

    Thanks for the formula but what do I do with the cos^-1 x? How do I figure out what x equals?

    Thanks
    Calculus Beginner
    In your question a = \frac{\sqrt{3}}{2} and you're trying to find the value of b ....
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