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

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

Mathematics
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Could it be simplified to (cos(x) + 1)^2?
and i am totally confused how to do it after that
Answers are 2pi, pi, pi/4 and 7pi/4, or pi/2 and 3pi/2

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Other answers:

your first step is correct, the next step is to take the square root of both sides
look at it like \(y^2+2y+1=0\)
so cos(x) = -1 ???
solve for y tell you found the values then replace y by cosx then solve
oh yes
yes! now find every x for which cos(x) = -1
Solve by putting y=cosx At last check values of y only in range of cosx then you will get the answer.
pi?
yup, good job!
sorry my computer froze for a bit
ty for your help
what if we extended the domain to (0, 5pi) what are the solutions ?
-1 + n2pi, n being a real integer from 0 to 2

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