How do you solve $2 {cos}^{2} x + cos x = 0$ $2cos^2x+cosx=0$ and find all exact general solutions?

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How do you solve $2 {cos}^{2} x + cos x = 0$ $2cos^2x+cosx=0$ and find all exact general solutions?

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Answer 1 · 2016-12-16T07:04:33

General Solution for $2 {cos}^{2} x + cos x = 0$ $2cos^2x+cosx=0$ is

$x = \frac{(2 n + 1) π}{2}$ $x=((2n+1)pi)/2$ or $x = 2 n π \pm \frac{2 π}{3}$ $x=2npi+-(2pi)/3$ , where $n$ $n$ is an integer.

Explanation:

$2 {cos}^{2} x + cos x = 0$ $2cos^2x+cosx=0$

$\Leftrightarrow cos x (2 cos x + 1) = 0$ $hArrcosx(2cosx+1)=0$

i.e. either $cos x = 0$ $cosx=0$ or $2 cos x + 1 = 0$ $2cosx+1=0$ i.e. $cos x = - \frac{1}{2}$ $cosx=-1/2$

General solution for $cos x = 0$ $cosx=0$ is $x = \frac{(2 n + 1) π}{2}$ $x=((2n+1)pi)/2$ , where $n$ $n$ is an integer

and general solution for $cos x = - \frac{1}{2} = cos (\pm \frac{2 π}{3})$ $cosx=-1/2=cos(+-(2pi)/3)$ is $x = 2 n π \pm \frac{2 π}{3}$ $x=2npi+-(2pi)/3$ , where $n$ $n$ is an integer

Hence, General Solution for $2 {cos}^{2} x + cos x = 0$ $2cos^2x+cosx=0$ is

$x = \frac{(2 n + 1) π}{2}$ $x=((2n+1)pi)/2$ or $x = 2 n π \pm \frac{2 π}{3}$ $x=2npi+-(2pi)/3$ , where $n$ $n$ is an integer.

Answer 2 · 2016-12-16T07:07:08

The solutions are $S = {\frac{π}{2} + 2 π n, \frac{3}{2} π + 2 π n, \frac{2}{3} π + 2 π n, \frac{4}{3} π + 2 π n}$ $S={pi/2+2pin,3/2pi+2pin,2/3pi+2pin,4/3pi+2pin}$

Explanation:

The equation is

$2 {cos}^{2} + cos x = 0$ $2cos^2+cosx=0$

$cos x (2 cos x + 1) = 0$ $cosx(2cosx+1)=0$

$cos x = 0$ $cosx=0$ and $cos x = - \frac{1}{2}$ $cosx=-1/2$

$cos x = 0$ $cosx=0$ gives $x = \frac{π}{2} + 2 π n$ $x=pi/2+2pin$ and $x = \frac{3}{2} π + 2 π n$ $x=3/2pi+2pin$

$cos x = - \frac{1}{2}$ $cosx=-1/2$ gives $x = \frac{2}{3} π + 2 π n$ $x=2/3pi+2pin$ and $x = \frac{4}{3} π + 2 π n$ $x=4/3pi+2pin$

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How do you solve $2 {cos}^{2} x + cos x = 0$ $2cos^2x+cosx=0$ and find all exact general solutions?

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