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Math Help - How do you do this? - Trigonometric Equation

  1. #1
    cdx
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    How do you do this? - Trigonometric Equation

    Solve

    <br />
sec^2 2x - 1, -2\pi \leqslant x \leqslant 2\pi<br />

    I'm getting weird answers for this question as opposed to the actual answer.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by cdx View Post
    Solve

    <br />
sec^2 2x - 1, -2\pi \leqslant x \leqslant 2\pi<br />

    I'm getting weird answers for this question as opposed to the actual answer.
    I assume you mean \sec^2(2x)-1=0

    This implies that either \sec(2x)=1 or \sec(2x)=-1

    Thus, find the solutions that fall in the given interval, such that x=\tfrac{1}{2}\sec^{-1}(-1) and x=\tfrac{1}{2}\sec^{-1}(1)

    Hint...look at the values of \cos^{-1}(-1) and \cos^{-1}(1)

    --Chris
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