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

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

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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)
User Rahul Shivsharan
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