2cos^2 x + 3cosx + 1 = 0

Solve on the interval {0,2Pi}

A. x = Pi/6, x = 7pi/6

B. x = Pi, x = 2pi/3, x = 4pi/3

C. x = 2pi, x = pi/3

D. x = 2pi, x = pi/4, x = 5pi/4

We will use the substitution:

u = cos x

2 u² + 3 u + 1 = 0

2 u² + u + 2 u + 1 = 0

u (2 u + 1) + (2 u + 1) = 0

(2 u + 1) (u + 1) = 0

2 u + 1 = 0, u = - 1/2 or: u + 1 = 0, u = - 1

cos x = - 1/2, x 1 = 2π/3, x 2 = 4π/3

cos x = - 1, x 3 = π

Answer: B)