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Find all solutions in the interval [0, 2π).

cos2x + 2 cos x + 1 = 0

1 Answer

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Answer:

x = pi/2, 3pi/2 x = pi

Explanation:

cos2x + 2 cos x + 1 = 0

cos 2x = cos^2 - sin^2x trig identity

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

rearrange

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

1 - sin^2 x = cos^2 x trig identity

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

combine like terms

2 cos ^2 x + 2 cos x = 0

divide by 2

cos ^2 x + cos x = 0

factor out a cos x

cos (x) (cos x +1) = 0

using the zero product property

cos (x) = 0 cos x + 1 =0

cos x =0 cos x = -1

taking the arccos of each side

arccos cos x = arccos (0) arccos (cos x) = arccos (-1)

x = pi/2, 3pi/2 x = pi



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