\(0≤θ≤2π \) (1)

\(3sinθ=sinθ -\sqrt{3}<=> 3sinθ-sinθ= -\sqrt{3}<=> 2sinθ= -\sqrt{3} <=>sinθ= -\frac{\sqrt{3}}{2}<=>sinθ=-sin(\frac{π}{3}),sinθ=-sin(\frac{2π}{3})<=>sinθ=sin(-\frac{π}{3}),sinθ=sin(-\frac{2π}{3})\)

With (1) \(sinθ=sin(2π-\frac{π}{3}),sinθ=sin(2π-\frac{2π}{3})<=>sinθ=sin(\frac{5π}{3}),sinθ=sin(\frac{4π}{3}) <=>θ=\frac{5π}{3},θ=\frac{4π}{3}\)

So \(θ=\frac{5π}{3},θ=\frac{4π}{3}\)

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