Question #dce98

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Question #dce98

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Answer 1 · 2016-10-15T00:48:30

Equation not true

Explanation:

Replace in the right side of the equation ${sin}^{2} x$ $sin^2 x$ by $(1 - {cos}^{2} x)$ $(1 - cos^2 x)$ , then solve the quadratic equation for cos x.
$cos x - 2 = 1 - {cos}^{2} x$ $cos x - 2 = 1 - cos^2 x$
${cos}^{2} x + cos x - 3 = 0$ $cos^2 x + cos x - 3 = 0$
Use the new improved quadratic formula (Socratic Search)
$D = d^{2} = b^{2} - 4 a c = 1 + 12 = 13$ $D = d^2 = b^2 - 4ac = 1 + 12 = 13$ --> $d = \pm 13$ $d = +- 13$
There are 2 real roots:
$cos x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{1}{2} \pm \frac{\sqrt{13}}{2} =$ $cos x = -b/(2a) +- d/(2a) = - 1/2 +- sqrt13/2 =$
Computing by calculator:
cos x = -0.5 + 1.80 = 1.30
cos x = - 0.5 - 1.80 = - 2.30
Both solutions are rejected (because -1 < x < 1). The equation is not true.

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Question #dce98

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