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Mathematics 8 Online
OpenStudy (anonymous):

Find all solutions to the equation. cos^2x + 2 cos x + 1 = 0

OpenStudy (luigi0210):

Factor the equation.

OpenStudy (anonymous):

is it cos=1/2

OpenStudy (luigi0210):

Nope try again

OpenStudy (anonymous):

is it cos - 1/2

OpenStudy (anonymous):

and then the solutions are 2pi/3 and 4pi/3

OpenStudy (jdoe0001):

notice is a quadratic equation, and you factor it just like any other quadratic \(\bf \Large cos^2x + 2 cos x + 1 = 0 \implies a^2+2a+1 = 0\)

OpenStudy (anonymous):

i still don't get it

OpenStudy (anonymous):

Can you write this ? \[\cos ^{2}x +2\cos x +1 = \left( \cos x + 1 \right)^{2}\]

OpenStudy (anonymous):

what do you mean?

OpenStudy (jdoe0001):

http://www.youtube.com/watch?v=eF6zYNzlZKQ

OpenStudy (ybarrap):

$$ \tt \Large{ \cos^2x+2\cos x + 1 =0\\ Let~u=\cos x\\ Then\\ u^2+2u+1=0\\ (u+1)(u+1)=0\\ (u+1)^2=0\\ (u+1)=0\\ u=-1\\ Sub~back~cos(x)~for~u\\ \cos x=-1\\ This~implies~that~x=-\pi ~\pm~2\pi n,~where~n\in\Bbb N. } $$ Let me know if you have any questions.

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