Respuesta :
sin2θ = 2sinθcosθ
2 sin 2θ − 3 sin θ = 0
2(2sinθcosθ) − 3 sin θ = 0
4sinθcosθ - 3sinθ = 0
sinθ(4cosθ - 3) = 0
sinθ = 0, 4cosθ - 3 = 0 cosθ = 3/4
θ = sin⁻¹(0) θ = cos⁻¹(0.75)
θ = 0, 180, 360 degrees, θ ≈ 41.41°, 360 - 41.41 = 318.59°
θ = 0°, 180°, 360°, ≈ 41.41°, ≈318.59°
2 sin 2θ − 3 sin θ = 0
2(2sinθcosθ) − 3 sin θ = 0
4sinθcosθ - 3sinθ = 0
sinθ(4cosθ - 3) = 0
sinθ = 0, 4cosθ - 3 = 0 cosθ = 3/4
θ = sin⁻¹(0) θ = cos⁻¹(0.75)
θ = 0, 180, 360 degrees, θ ≈ 41.41°, 360 - 41.41 = 318.59°
θ = 0°, 180°, 360°, ≈ 41.41°, ≈318.59°