TylerRaymond6205 TylerRaymond6205 22-02-2024 Mathematics contestada 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
