Question #f6450

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Question #f6450

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Answer 1 · 2017-11-28T22:17:34

I am assuming that we are solving for,

$4 {cos}^{2} θ + 4 cos θ + 1 = 0$ $4cos^2 theta +4cos theta +1 = 0$

This is a quadratic polynomial of the form;

$a x^{2} + b x + c = 0$ $ax^2 + bx + c = 0$

Where $x = cos θ$ $x=cos theta$ . To start, lets get rid of the 4 in the first term. We can divide both sides of the equation by 4 to get;

${cos}^{2} θ + cos θ + \frac{1}{4} = 0$ $cos^2 theta + cos theta +1/4 = 0$

There are several methods for solving quadratics, but at a glance, I notice that $2 \times \frac{1}{2} = 1$ $2 xx 1/2 = 1$ and ${(\frac{1}{2})}^{2} = \frac{1}{4}$ $(1/2)^2 = 1/4$ , which are the last two coefficients in our polynomial. Therefore, we can rewrite the quadratic as;

${(cos θ + \frac{1}{2})}^{2} = 0$ $(cos theta +1/2)^2 = 0$

Taking the square root of both sides, we have;

$cos θ + \frac{1}{2} = 0$ $cos theta + 1/2 = 0$

So;

$cos θ = - \frac{1}{2}$ $cos theta = -1/2$

Plugging ${cos}^{- 1} (- \frac{1}{2})$ $cos^-1(-1/2)$ into a calculator, or checking a unit circle will reveal that;

$θ = 120^{\circ}, 240^{\circ}$ $theta = 120^@, 240^@$

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Question #f6450

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