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

cos2x + 2 cos x + 1 = 0

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

Step-by-step 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