How do you solve ${cos}^{2} x - 4 cos x = 1$ $cos^2 x - 4 cos x = 1$ over the interval 0 to 2pi?

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How do you solve ${cos}^{2} x - 4 cos x = 1$ $cos^2 x - 4 cos x = 1$ over the interval 0 to 2pi?

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Answer 1 · 2016-04-05T00:28:15

$103^{\circ} 65 and 256^{\circ} 35$ $103^@65 and 256^@35$

Explanation:

$f (x) = {cos}^{2} x - 4 cos x - 1 = 0$ $f(x) = cos^2 x - 4cos x - 1 = 0$
$D = d^{2} = b^{2} - 4 a c = 16 + 4 = 20$ $D = d^2 = b^2 - 4ac = 16 + 4 = 20$ ---> $d = \pm 2 \sqrt{5}$ $d = +- 2sqrt5$
There are 2 real roots.
$cos x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{4}{2} \pm 2 \frac{\sqrt{5}}{2} = 2 \pm \sqrt{5}$ $cos x = -b/(2a) +- d/(2a) = 4/2 +- 2sqrt5/2 = 2 +- sqrt5$

a . $cos x = 2 + \sqrt{5}$ $cos x = 2 + sqrt5$ (Rejected because > 1)
b. $cos x = 2 - \sqrt{5} = 2 - 2.24 = - 0.24$ $cos x = 2 - sqrt5 = 2 - 2.24 = - 0. 24$ --> $x = \pm 103^{\circ} 65$ $x = +- 103^@65$
Answer for $(0, 2 π) :$ $(0, 2pi):$
$103^{\circ} 65 and 256^{\circ} 35$ $103^@65 and 256^@35$ (co-terminal to - 103.65)

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How do you solve ${cos}^{2} x - 4 cos x = 1$ $cos^2 x - 4 cos x = 1$ over the interval 0 to 2pi?

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How do you solve ${cos}^{2} x - 4 cos x = 1$ $cos^2 x - 4 cos x = 1$ over the interval 0 to 2pi?