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Thread: Need help with trigonometric equations please.

  1. #1
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    Need help with trigonometric equations please.

    How would I solve the trig equation 5 cos(2x)=2 ?????????????
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  2. #2
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    Re: Need help with trigonometric equations please.

    $\displaystyle \cos 2x = \frac{2}{5}$

    What angles within the unit circle have a cosine of 2/5?
    Last edited by richard1234; Nov 11th 2012 at 09:14 AM.
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  3. #3
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    Re: Need help with trigonometric equations please.

    assuming solutions in the interval $\displaystyle 0 < x < 2\pi$ ...


    $\displaystyle 0 < x < 2\pi \implies 0 < 2x < 4\pi$

    let $\displaystyle t = 2x$ , so $\displaystyle 0 < t < 4\pi$


    $\displaystyle 5 \cos{t} = 2$

    $\displaystyle \cos{t} = \frac{2}{5}$

    since $\displaystyle \cos{t} > 0$ , $\displaystyle t$ is an angle in quads I or IV


    quad I solutions for $\displaystyle t$ , (two) ...

    $\displaystyle t = \arccos\left(\frac{2}{5}\right)$ and $\displaystyle t = 2\pi + \arccos\left(\frac{2}{5}\right)$

    so $\displaystyle x = \frac{1}{2}\arccos\left(\frac{2}{5}\right)$ and $\displaystyle x = \pi + \frac{1}{2}\arccos\left(\frac{2}{5}\right)$


    quad IV solutions (two) ...

    $\displaystyle t = 2\pi - \arccos\left(\frac{2}{5}\right)$ and $\displaystyle t = 4\pi - \arccos\left(\frac{2}{5}\right)$

    so $\displaystyle x = \pi - \frac{1}{2}\arccos\left(\frac{2}{5}\right)$ and $\displaystyle x = 2\pi - \frac{1}{2}\arccos\left(\frac{2}{5}\right)$
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