What values for theta (0<=theta<=2pi) satify the equation?

\(0≤θ≤2π \) (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 \(θ=\frac{5π}{3},θ=\frac{4π}{3}\)

\(3\sin\theta\ =\ \sin\theta-\sqrt3\\~\\ 3\sin\theta-\sin\theta\ =\ -\sqrt3\\~\\ 2\sin\theta\ =\ -\sqrt3\\~\\ \sin\theta\ =\ -\frac{\sqrt3}{2}\)

We can look at a unit circle to see what values of  θ  in the interval  [0, 2π]  will satisfy  \(\sin\theta=-\frac{\sqrt3}{2}\) .

Here is a pretty clear unit circle:

https://en.wikibooks.org/wiki/Trigonometry/Trigonometric_Unit_Circle_and_Graph_Reference

\(\theta\ =\ \frac{4\pi}{3}\\~\\ \theta\ =\ \frac{5\pi}{3}\)