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

cos2x + 2 cos x + 1 = 0

User Zuuum
by
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1 Answer

3 votes

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



User Cattode
by
6.3k points