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Mathematics 21 Online
OpenStudy (anonymous):

Find all solutions in the interval [0, 2π). cos2x + 2 cos x + 1 = 0

OpenStudy (anonymous):

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

OpenStudy (cwrw238):

this is a quadratic equation which will factorise

OpenStudy (anonymous):

\[\cos²(x) + 2\cos(x) + 1 = (\cos(x) + 1)²\] Hence the roots are all x out of [0,2pi) for which cos(x) = -1. -pi and pi

OpenStudy (anonymous):

Sorry, of course it's just + pi as there are no negative numbers in this interval

OpenStudy (cwrw238):

what about 0 as a solution?

OpenStudy (cwrw238):

no sorry - its -1

OpenStudy (cwrw238):

cos 0 = 1 of course

OpenStudy (anonymous):

the answer choices are a) x=2pi B)x=pi c) x=pi/4, 7pi/4

OpenStudy (cwrw238):

are you sure you have the correct question is it cos 2x or cos^2 x ?

OpenStudy (anonymous):

it is cos^2 x

OpenStudy (anonymous):

It's B) x = pi as in my answer.

OpenStudy (cwrw238):

in that case the answer is as Stiwam said

OpenStudy (anonymous):

thankyou!

OpenStudy (cwrw238):

yw

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