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Math Help - Analytic Trig

  1. #1
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    Analytic Trig

    Solve each equation on the interval 0 greater than or equal to theta or less than 2pie.

    12) 4 (cos^2) theta - 3 = 0




    Please be descriptive I have having difficulties with these types. TY
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  2. #2
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    Quote Originally Posted by kukid123 View Post
    Solve each equation on the interval 0 greater than or equal to theta or less than 2pie.

    12) 4 (cos^2) theta - 3 = 0




    Please be descriptive I have having difficulties with these types. TY
    Just "unwrap" it like you would any other equation.

    4\cos^2{\theta} - 3 = 0 Add 3 to both sides.

    4\cos^2{\theta} = 3 Divide by 4.

    \cos^2{\theta} = \frac{3}{4}. Take the square root.

    \cos{\theta} = \pm\frac{\sqrt{3}}{2}.


    So what values of \theta make \cos{\theta} = \frac{\sqrt{3}}{2} or -\frac{\sqrt{3}}{2}.


    Since the answer is both positive and negative, there are answers in all quadrants. Let's start with the focus angle.

    We know \cos{\frac{\pi}{6}} = \frac{\sqrt{3}}{2}

    So \frac{\pi}{6} is the focus angle.

    In the domain 0 \leq x \leq 2\pi the solutions are...

    x = \left\{\frac{\pi}{6}, \pi - \frac{\pi}{6}, \pi + \frac{\pi}{6}, 2\pi - \frac{\pi}{6}\right\}

     = \left\{\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}\right\}.


    Does that make sense?
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  3. #3
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    Yeah thanks bro. Great work.
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