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Math Help - need help with a trig problem

  1. #1
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    need help with a trig problem

    this is the problem: 2 csc²x=3 cot²x-1
    this is what I have so far: see attached file
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  2. #2
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    Quote Originally Posted by RaphaelB30 View Post
    this is the problem: 2 csc²x=3 cot²x-1
    this is what I have so far: see attached file
    \frac{2}{sin^2(x)} = \frac{3cos^2(x)}{sin^2(x)} - 1

    Getting the same denominator on the right:

    \frac{3cos^2(x) - sin^2(x)}{sin^2(x)}

    Now sin^2(x) will cancel

    2 = 3cos^2(x) - sin^2(x)

    2 = 3cos^2(x) - (1-cos^2(x)) = 3cos^2(x) - 1 + cos^2(x) = 4cos^2(x) - 1

    4cos^2(x) -3 = 0

    using the difference of two squares:

    (2cos(x) - \sqrt3)(2cos(x) + \sqrt3) = 0

    so cos(x) = ±\frac{\sqrt3}{2} and thus

    x = arcos(\frac{\sqrt3}{2}) = \frac{5\pi}{6} or

    x = arcos(-\frac{\sqrt3}{2}) = \frac{\pi}{6}
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  3. #3
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    2 csc²x=3 cot²x-1
    2(\cot^2{x} + 1) = 3\cot^2{x} - 1

    0 = \cot^2{x} - 3

    0 = (\cot{x} - \sqrt{3})(\cot{x} + \sqrt{3})

    \cot{x} = \sqrt{3}

    x = \frac{\pi}{6} , x = \frac{7\pi}{6}

    \cot{x} = -\sqrt{3}

    x = \frac{5\pi}{6} , x = \frac{11\pi}{6}
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