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Thread: Solving a trigonometric equation

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    Solving a trigonometric equation

    5 Cos ^2θ – 2=4 Cos θ
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    Quote Originally Posted by blame_canada100 View Post
    5 Cos ^2θ 2=4 Cos θ
    note that this is the same as $\displaystyle 5 \cos^2 \theta - 4 \cos \theta - 2 = 0$.

    now, note that this is a quadratic equation in $\displaystyle \cos \theta$. to see this easier, let $\displaystyle y = \cos \theta$, then you have

    $\displaystyle 5y^2 - 4y - 2 = 0$

    now solve for $\displaystyle y$, and then set it equal to $\displaystyle \cos \theta$ to solve for $\displaystyle \theta$
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    Quote Originally Posted by blame_canada100 View Post
    5 Cos ^2θ 2=4 Cos θ
    You have a quadratic equation.

    Let $\displaystyle X = \cos{\theta}$.

    So the equation becomes

    $\displaystyle 5X^2 - 2 = 4X$

    $\displaystyle 5X^2 - 4X - 2 = 0$.

    Solving for X using the Quadratic formula gives

    $\displaystyle X = \frac{2 + \sqrt{14}}{5}$ and $\displaystyle X = \frac{2 - \sqrt{14}}{5}$.

    Since $\displaystyle X = \cos{\theta}$, this gives

    $\displaystyle \cos{\theta} = \frac{2 + \sqrt{14}}{5}$ or $\displaystyle \cos{\theta} = \frac{2 - \sqrt{14}}{5}$.

    Remember that $\displaystyle -1\leq \cos{\theta} \leq 1$.

    Only one of the above equations can satisfy this. Solve for $\displaystyle \theta$ in the appropriate equation.
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