Solve the equation on the interval [0, 2pi)

2 cos^2 (theta/2) - 3 cos theta = 0

Hello:

by idetity : 1+cos (2a) = 2cos² (a)

let : a = θ/2

1+cos (2 * θ/2) = 2cos² (θ/2)

1+cos (θ) = 2cos² (θ/2)

solve : 2 cos^2 (θ/2) - 3 cos (θ) = 0 ... (1)

(1) : 1+cos (θ) - 3 cos (θ) = 0

2cos (θ) = 1

cos (θ) = 1/2

in : [0, 2π [

(θ = π/3) or (θ = 5π/3)