How do you solve $2 {cos}^{2} x + 3 cos x - 2 = 0$ $2cos^2x+3cosx-2=0$ ?

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How do you solve $2 {cos}^{2} x + 3 cos x - 2 = 0$ $2cos^2x+3cosx-2=0$ ?

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Answer 1 · 2016-11-27T23:11:28

Let $n = cos x$ $n=cosx$

$2 n^{2} + 3 n - 2 = 0$ $2n^2+3n-2=0$

Solve for n by factoring:
$2 n^{2} - n + 4 n - 2 = 0$ $2n^2-n+4n-2=0$
$n (2 n - 1) + 2 (2 n - 1) = 0$ $n(2n-1)+2(2n-1)=0$
$(n + 2) (2 n - 1) = 0$ $(n+2)(2n-1)=0$

$n + 2 = 0$ $n+2=0$
$n = - 2$ $n=-2$

$2 n - 1 = 0$ $2n-1=0$
$n = \frac{1}{2}$ $n=1/2$

Substitute $n = cos x$ $n=cosx$ in for those two answers:
$cosxcancel(=)-2$
Cosine can never equal -2.

$cosx=1/2$

$x=pi/3, (5pi)/3$

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How do you solve $2 {cos}^{2} x + 3 cos x - 2 = 0$ $2cos^2x+3cosx-2=0$ ?

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How do you solve $2 {cos}^{2} x + 3 cos x - 2 = 0$ $2cos^2x+3cosx-2=0$ ?