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Thread: need help with a trig problem

  1. #1
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    need help with a trig problem

    this is the problem: 2 cscx=3 cotx-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 cscx=3 cotx-1
    this is what I have so far: see attached file
    $\displaystyle \frac{2}{sin^2(x)} = \frac{3cos^2(x)}{sin^2(x)} - 1$

    Getting the same denominator on the right:

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

    Now $\displaystyle sin^2(x)$ will cancel

    $\displaystyle 2 = 3cos^2(x) - sin^2(x)$

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

    $\displaystyle 4cos^2(x) -3 = 0$

    using the difference of two squares:

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

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

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

    $\displaystyle x = arcos(-\frac{\sqrt3}{2}) = \frac{\pi}{6} $
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  3. #3
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    2 cscx=3 cotx-1
    $\displaystyle 2(\cot^2{x} + 1) = 3\cot^2{x} - 1$

    $\displaystyle 0 = \cot^2{x} - 3$

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

    $\displaystyle \cot{x} = \sqrt{3}$

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

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

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