$\begin{aligned} & 2 \sin ^3 x+2 \sin x \cdot \cos ^2 x+4 \sin x-4=0 \\ & 2 \sin ^3 x+2 \sin x \cdot\left(1-\sin ^2 x\right)+4 \sin x-4=0 \\ & 6 \sin x-4=0 \\ & \sin x=\frac{2}{3} \\ & \mathbf{n}=5 \text { (in the given interval) } \\ & x^2+5 x+2=0 \\ & x=\frac{-5 \pm \sqrt{17}}{2} \\ & \text { Required interval }(-\infty, 0) \end{aligned}$