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Problem Find the general solution of sin 2 θ = 3 2 \sin 2\theta = \dfrac{\sqrt3}{2} sin 2 θ = 2 3 θ = π 3 + n π or θ = 2 π 3 + n π , n ∈ Z \theta = \dfrac{\pi}{3}+n\pi \;\text{or}\; \theta = \dfrac{2\pi}{3}+n\pi,\; n\in\mathbb Z θ = 3 π + nπ or θ = 3 2 π + nπ , n ∈ Z θ = π 6 + n π or θ = π 3 + n π , n ∈ Z \theta = \dfrac{\pi}{6}+n\pi \;\text{or}\; \theta = \dfrac{\pi}{3}+n\pi,\; n\in\mathbb Z θ = 6 π + nπ or θ = 3 π + nπ , n ∈ Z θ = π 6 + 2 n π or θ = 5 π 6 + 2 n π , n ∈ Z \theta = \dfrac{\pi}{6}+2n\pi \;\text{or}\; \theta = \dfrac{5\pi}{6}+2n\pi,\; n\in\mathbb Z θ = 6 π + 2 nπ or θ = 6 5 π + 2 nπ , n ∈ Z θ = π 12 + n π or θ = 5 π 12 + n π , n ∈ Z \theta = \dfrac{\pi}{12}+n\pi \;\text{or}\; \theta = \dfrac{5\pi}{12}+n\pi,\; n\in\mathbb Z θ = 12 π + nπ or θ = 12 5 π + nπ , n ∈ Z Browse the problem library