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Mathematics 25 Online
OpenStudy (anonymous):

Find all solutions to the equation. cos2x + 2 cos x + 1 = 0

OpenStudy (anonymous):

ambiguous!

OpenStudy (math&ing001):

Does cos2x mean \[\cos ^{2}x ?\]

OpenStudy (mertsj):

Factor the left side.

OpenStudy (anonymous):

cos2x = cosx^2 - sinx^2 and 1=sinx^2 + cosx^2. cancel the sinx^2 and you get 2cosx^2 + 2cosx = 0 set cosx = u and then factor it or use quadratic formula.

OpenStudy (mertsj):

@melacho If you want help you MUST clarify: Is it cos(2x) or cos^2(x) ???

OpenStudy (anonymous):

and is there a domain?

OpenStudy (anonymous):

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

OpenStudy (mertsj):

Then factor the left side. It is like this: x^2+2x+1

OpenStudy (jdoe0001):

notice is just a quadratic, thus you'd solve it like you'd any quadratic

OpenStudy (anonymous):

I am so confused >.< I'm sorry

OpenStudy (anonymous):

(cosx + 1)^2 = 0 better?

OpenStudy (anonymous):

cos(x)+1=0 cos(x)=1 x 3pi/2kpi

OpenStudy (mertsj):

cos x = -1 x=pi + 2kpi

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