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Math Help - trigonometric functions

  1. #1
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    trigonometric functions

    I've been working on this problem for awhile now but I think it's easier than I'm making it seem. Here's what I have so far:

    3sec^2x = 4
    sec^2x = 4/3
    secx = sqrt 4/3

    It seems right to me but of course sqrt 4/3 isn't a proper answer, since it's not within the domain of sec x. Am I doing this right, and is the answer 'undefined,' or is there something wrong? Please help me, thanks.
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  2. #2
    o_O
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     \sec^{2} x = \frac{4}{3}
    \cos^{2} x = \frac{3}{4} (by taking the reciprical)
    \cos x = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{\sqrt{4}} = \pm \frac{\sqrt{3}}{2}



    I'm not sure what made you lead to the conclusion that your answer leads to sec x being undefined ..
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  3. #3
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    Quote Originally Posted by o_O View Post
     \sec^{2} x = \frac{4}{3}
    \cos^{2} x = \frac{3}{4} (by taking the reciprical)
    \cos x = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{\sqrt{4}} = \pm \frac{\sqrt{3}}{2}



    I'm not sure what made you lead to the conclusion that your answer leads to sec x being undefined ..
    Ahh, thank you so much! I put it in my calculator and it gave me an error. My teacher's right, then, when she says we shouldn't depend on them.
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