Question #34897

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Question #34897

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Answer 1 · 2017-03-19T18:03:20

$\pm 68^{\circ} 28 + k 360^{\circ}$ $+- 68^@28 + k360^@$

Explanation:

Solve this quadratic equation for cos x by using the improved quadratic formula (Socratic Search)
$2 {cos}^{2} x + 2 cos x - 1 = 0$ $2cos^2 x + 2cos x - 1 = 0$
$D = d^{2} = b^{2} - 4 a c = 4 + 8 = 12$ $D = d^2 = b^2 - 4ac = 4 + 8 = 12$ --> $d = \pm 2 \sqrt{3}$ $d = +- 2sqrt3$
There are 2 real roots:
$cos x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{1}{2} \pm \frac{\sqrt{3}}{2} = \frac{- 1 \pm \sqrt{3}}{2}$ $cos x = -b/(2a) +- d/(2a) = - 1/2 +- sqrt3/2 = (-1 +- sqrt3)/2$
2 solutions for cos x:
$cos x = \frac{- 1 + \sqrt{3}}{2} = \frac{0.73}{2} = 0.37$ $cos x = (-1 + sqrt3)/2 = 0.73/2 = 0.37$ , and
$cos x = \frac{- 1 - \sqrt{3}}{2} = - \frac{2.73}{2} = - 1.37$ $cos x = (- 1 - sqrt3)/2 = -2.73/2 = - 1.37$ (Rejected as < - 1)
Use unit circle and calculator:
$cos x = 0.37$ $cos x = 0.37$ --> $x = \pm 68^{\circ} 28 + k 360^{\circ}$ $x = +- 68^@28 + k360^@$

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Question #34897

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