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Math Help - Trigonometry Help!

  1. #1
    Member looi76's Avatar
    Joined
    Jan 2008
    Posts
    185

    Trigonometry Help!

    Question:
    Find the solution of \sin{x}-1=0 if 0 \leq x <  \frac{1}{2}\pi.

    Attempt:

    \sin{x}-1=0

    \sin{x} = 1

    x = \sin^{-1}{(1)}

    x = 90^o

    x = \frac{1}{2}\pi

    Either my answer is wrong or the question is wrong because it was mentioned in the question that it is not greater than \frac{1}{2}\pi. Am I correct?
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  2. #2
    Senior Member
    Joined
    Nov 2007
    Posts
    329
    You're correct, so the answer is that no real number solves this equation on the given interval.
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