Question

2 sin 2θ − 3 sin θ = 0

Answer 1

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°

Answer 2

(-1/4)
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