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Math Help - 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 5 \cos^2 \theta - 4 \cos \theta - 2 = 0.

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

    5y^2 - 4y - 2 = 0

    now solve for y, and then set it equal to \cos \theta to solve for \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 X = \cos{\theta}.

    So the equation becomes

    5X^2 - 2 = 4X

    5X^2 - 4X - 2 = 0.

    Solving for X using the Quadratic formula gives

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

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

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

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

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