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

1010: 3030

is it i think