How do you solve $2sin^2x - cosx = 1$ over the interval 0 to 2pi?

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Answer 1 · 2016-02-14T03:55:27

$x=60^@, 300^@, 180^@" "$ or
$x=pi/3, (5pi)/3, pi$

Start from the given $2 sin^2 x - cos x = 1$

note that, $sin^2 x+cos^2 x=1$ and $sin^2 x=1-cos^2 x$

and $2 sin^2 x - cos x = 1$

is also $2*(1-cos^2 x)-cos x=1$

and

$2-2cos^2 x-cos x=1$

$2 cos^2 x + cos x-2+1=0$

$2 cos^2 x + cos x-1=0$

use $color(red)("Quadratic Equation")$

$cos x=(-b+-sqrt(b^2-4ac))/(2a)$

$2 cos^2 x + cos x-1=0$

from $2 cos^2 x + 1*cos x-1=0$

Let $a=2$ , $b=1$ , $c=-1$

$cos x=(-b+-sqrt(b^2-4ac))/(2a)$

$cos x=(-1+-sqrt(1^2-4*(2)*(-1)))/(2*2)$

$cos x=(-1+-3)/4$

$cos x_1=1/2$ and $cos x_2=-1$

$x_1=60^@, 300^@$
$x_2=180^@$

God bless America ....

How do you solve 2sin^2x - cosx = 1 2sin^2x - cosx = 1 over the interval 0 to 2pi?