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

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Answer 1 · 2016-12-16T07:04:33

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

$x=((2n+1)pi)/2$ or $x=2npi+-(2pi)/3$ , where $n$ is an integer.

$2cos^2x+cosx=0$

$hArrcosx(2cosx+1)=0$

i.e. either $cosx=0$ or $2cosx+1=0$ i.e. $cosx=-1/2$

General solution for $cosx=0$ is $x=((2n+1)pi)/2$ , where $n$ is an integer

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

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

$x=((2n+1)pi)/2$ or $x=2npi+-(2pi)/3$ , where $n$ is an integer.

Answer 2 · 2016-12-16T07:07:08

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

The equation is

$2cos^2+cosx=0$

$cosx(2cosx+1)=0$

$cosx=0$ and $cosx=-1/2$

$cosx=0$ gives $x=pi/2+2pin$ and $x=3/2pi+2pin$

$cosx=-1/2$ gives $x=2/3pi+2pin$ and $x=4/3pi+2pin$

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