Find $0^\circ < x < 360^\circ$

Using half-angle tangent substitution , we can solve the problem as follows:

$\begin{aligned} 3 \sin x & = (1-\cos x)(3-\cos x) & \small \blue{\text{Let }t = \tan \frac x2} \\ 3\times \frac {2t}{1+t^2} & = \left(1-\frac {1-t^2}{1+t^2} \right)\left(3-\frac {1-t^2}{1+t^2} \right) \\ 6t(1+t^2) & = 2t^2(2+4t^2) \\ 3 + 3t^2 & = 2t + 4t^3 \\ 4t^3 - 3t^2 + 2t - 3 & = 0 \\ (t-1)(4t^2+t+3) & = 0 \\ \implies t = \tan \frac x2 & = 1 & \small \blue{\text{Only real solution}} \\ \implies x & = \boxed{90^\circ} & \small \blue{\text{for }0^\circ \le x < 360^\circ} \end{aligned}$