The solution of the equation 2sin(θ) + 1 = 0 is θ = 7π/6 + 2nπ and θ = 11π/6 + 2nπ, where n is an integer.

A) θ = 5π/6 + 2nπ and θ = π/6 + 2nπ
B) θ = 5π/6 + nπ and θ = π/6 + nπ
C) θ = π/6 + 2nπ and θ = 5π/6 + 2nπ
D) θ = π/6 + nπ and θ = 5π/6 + nπ