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

$\theta\ =\ \frac{4\pi}{3}\\~\\ \theta\ =\ \frac{5\pi}{3}$