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Math Help - Find x, anyone help?

  1. #1
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    Find x, anyone help?

    sinx - x multiplied by cosx - (pi/2)=0

    please note, working in radians

    any method to give the solution thats not an approximation

    please email with any solutions

    thanks
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  2. #2
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    Quote Originally Posted by LaurenB View Post
    sinx - x multiplied by cosx - (pi/2)=0

    please note, working in radians

    any method to give the solution thats not an approximation

    please email with any solutions

    thanks
    Well the method I propose uses routine algebra and recognition of special angles and values.

    Using the null factor law (you might know it by a different name):

    \sin (x) - x = 0 \Rightarrow \sin x = x .... (1)

    or

    \cos (x) - \frac{\pi}{2} = 0 \Rightarrow \cos x = \frac{\pi}{2} .... (2)

    Regarding equation (1), an obvious exact solution is x = 0. The tricky bit is to show that this actually the only solution ......



    Edit: Deleted my remarks on equation (2) for reasons made obvious by an ancient Greek philosopher!
    Last edited by mr fantastic; May 10th 2008 at 03:41 PM.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    \cos (x) - \frac{\pi}{2} = 0 \Rightarrow \cos x = \frac{\pi}{2} .... (2)
    Equation (2) is obviously your best chance for exact solutions: x = \pm \frac{\pi}{2}, ~ \pm \frac{3\pi}{2}, ~ \pm \frac{5\pi}{2} ...... and in general x = \frac{(2n-1) \pi}{2} where n is any integer.
    Do you want to rethink that answer?
    Maybe \arccos \left( {\frac{\pi }{2}} \right)?
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  4. #4
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    Quote Originally Posted by Plato View Post
    Do you want to rethink that answer?
    Maybe \arccos \left( {\frac{\pi }{2}} \right)?


    Thankyou.

    And \frac{\pi}{2} > 1 leads to the obvious conclusion .......
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