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Thread: 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.

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

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

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

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


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


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

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

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

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

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

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


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