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mgarcia634
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Find all solutions in the interval [0, 2π).
cos2x + 2 cos x + 1 = 0
 one year ago
 one year ago
mgarcia634 Group Title
Find all solutions in the interval [0, 2π). cos2x + 2 cos x + 1 = 0
 one year ago
 one year ago

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mgarcia634 Group TitleBest ResponseYou've already chosen the best response.0
cos^2 x+ 2 cos x +1=0
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
this is a quadratic equation which will factorise
 one year ago

Stiwan Group TitleBest ResponseYou've already chosen the best response.1
\[\cos²(x) + 2\cos(x) + 1 = (\cos(x) + 1)²\] Hence the roots are all x out of [0,2pi) for which cos(x) = 1. pi and pi
 one year ago

Stiwan Group TitleBest ResponseYou've already chosen the best response.1
Sorry, of course it's just + pi as there are no negative numbers in this interval
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
what about 0 as a solution?
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
no sorry  its 1
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
cos 0 = 1 of course
 one year ago

mgarcia634 Group TitleBest ResponseYou've already chosen the best response.0
the answer choices are a) x=2pi B)x=pi c) x=pi/4, 7pi/4
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
are you sure you have the correct question is it cos 2x or cos^2 x ?
 one year ago

mgarcia634 Group TitleBest ResponseYou've already chosen the best response.0
it is cos^2 x
 one year ago

Stiwan Group TitleBest ResponseYou've already chosen the best response.1
It's B) x = pi as in my answer.
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
in that case the answer is as Stiwam said
 one year ago

mgarcia634 Group TitleBest ResponseYou've already chosen the best response.0
thankyou!
 one year ago
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