How do you use the quadratic formula to solve $4 {cos}^{2} x - 4 cos x - 1 = 0$ $4cos^2x-4cosx-1=0$ in the interval $[0, 2 π)$ $[0,2pi)$ ?

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How do you use the quadratic formula to solve $4 {cos}^{2} x - 4 cos x - 1 = 0$ $4cos^2x-4cosx-1=0$ in the interval $[0, 2 π)$ $[0,2pi)$ ?

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Answer 1 · 2017-02-17T01:31:42

$101^{\circ} 95; 258^{\circ} 05$ $101^@95 ; 258^@05$

Explanation:

Solve this quadratic equation for cos x by using the improved quadratic formula:
$f (x) = 4 {cos}^{2} x - 4 cos x - 1 = 0$ $f(x) = 4cos^2 x - 4cosx - 1 = 0$
$D = d^{2} = b^{2} - 4 a c = 16 + 16 = 32$ $D = d^2 = b^2 - 4ac = 16 + 16 = 32$ --> $d = \pm 4 \sqrt{2}$ $d = +- 4sqrt2$
There are 2 real roots:
$cos x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{4}{8} \pm 4 \frac{\sqrt{2}}{8} = \frac{1 \pm \sqrt{2}}{2}$ $cos x = - b/(2a) +- d/(2a) = 4/8 +- 4sqrt2/8 = (1 +- sqrt2)/2$
$cos x 1 = \frac{1 + \sqrt{2}}{2} = \frac{2.414}{2} = 1.207$ $cos x1 = (1 + sqrt2)/2 = 2.414/2 = 1.207$ (Rejected as > 1)
$cos x 2 = \frac{1 - \sqrt{2}}{2} = - \frac{0.414}{2} = - 0.207$ $cos x2 = (1 - sqrt2)/2 = - 0.414/2 = - 0.207$
cos x2 = - 0.207 --> $x = \pm 101^{\circ} 95$ $x = +- 101^@95$
x = 258.05 is co-terminal to (- 101.95)
Answers for $(0, 2 π)$ $(0, 2pi)$
$101^{\circ} 95; 258^{\circ} 05$ $101^@95 ; 258^@05$

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How do you use the quadratic formula to solve $4 {cos}^{2} x - 4 cos x - 1 = 0$ $4cos^2x-4cosx-1=0$ in the interval $[0, 2 π)$ $[0,2pi)$ ?

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Explanation:

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How do you use the quadratic formula to solve $4 {cos}^{2} x - 4 cos x - 1 = 0$ $4cos^2x-4cosx-1=0$ in the interval $[0, 2 π)$ $[0,2pi)$ ?