I need help on this problem please. Find exact values of cos(2theta)=sqrt3/2. In [0,2pi). Please.?

It looks like you limited your solutions too soon. [0,2pi) is the interval for x, not 2x. (Using x instead of theta.)

cos 2x = √3 / 2
2x = ± [cos⁻¹ (√3 / 2)] + 2nπ ... full general solution, n is any integer
2x = ±π/6 + 2nπ
x = ±π/12 + nπ

The x values in [0,2π] are π/12, 13π/12 for the + case, and 11π/12, 23π/12 for the - case.

θ is in [0, 2π)
2θ is in [0, 4π)

cos(2θ) = (√3)/2
2θ = π/6, 2π - (π/6), 2π + (π/6), 4π - (π/6)
2θ = π/6, 11π/6, 13π/6, 23π/6
θ = π/12, 11π/12, 13π/12, 23π/12

　
Since 0 ≤ θ < 2π, then 0 ≤ 2θ < 4π

cos 2θ = √3/2
2θ = π/6, 11π/6, 13π/6, 23π/6
θ = π/12, 11π/12, 13π/12, 23π/12