Find all solutions to the equation.

Question

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

Answer 1

Given cos^2x + 2cosx + 1 = 0

cos^2x + cosx + cosx + 1 = 0

Taking cosx common from first two terms and 1 from last two terms

cosx(cosx + 1) + 1 ( cosx + 1) = 0

(cosx + 1)(cosx + 1)= 0

cosx + 1 = 0 and cosx + 1 = 0

cosx = -1 and cosx = -1

x = cos^ -1(-1) and x = cos^-1(-1)

x = 180 and x = 180

Therefore total solutions is (180, 180)

Answer 2

cos^2x + 2cosx + 1 = 0

Recall : a^2 + b^2 + 2ab = (a + b)^2

(cosx + 1)^2 = 0

cosx + 1 = 0 or cosx + 1 = 0

cosx = -1 or cosx = -1

x = pi, 3pi, 5pi , 7pi , 9pi ..................

The solutions of the equation are pi, 3pi , 5pi ,.............

Comments

cos (x) = - 1.

cos (x) = cos(π)

The genaral solution of cos(θ) = cos(α) is θ = 2nπ ± α, where n is an integer.

⇒ x = 2nπ ± π.

Therefore, the solution is x = 2nπ ± π, where n is an integer.