How do you solve $2 {cos}^{2} x + cos x - 1 = 0$ $2cos^2x + cosx - 1= 0$ and find the general solution?

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Answer 1 · 2016-04-23T09:04:42

$x = 2kpi+-pi/3, k+0, +-1, +-2, ...$ and
x=an odd multiple of $pi=(2k+1)pi, k=0. +-1, +-2...$

This is a quadratic in cos x.

The roots are
$cos x =1/2 and -1$

The corresponding principal values of
arc cos (1/2) and arc cos(-1) are
$pi/3 and pi$ .

The general values are

$x = 2kpi+-pi/3, k+0, +-1, +-2, ...$ and

x=an odd multiple of $pi=(2k+1)pi, k=0. +-1, +-2...$

For the first root, there is a choice from $+-pi/3 or pi/3 and (5pi)/3$ .

For the principal value. For the second, there is no choice.,

How do you solve 2cos^2x + cosx - 1= 02cos2x+cosx−1=02cos^2x + cosx - 1= 0 and find the general solution?