To find the solutions of the equation $\sqrt{3} \operatorname{cosec}^2 \theta - 2(\sqrt{3} - 1) \operatorname{cosec} \theta - 4 = 0$, we can begin by solving for $\operatorname{cosec} \theta$. The quadratic equation in terms of $\operatorname{cosec} \theta$ is: $\sqrt{3} \operatorname{cosec}^2 \theta - 2(\sqrt{3} - 1) \operatorname{cosec} \theta - 4 = 0$ Using the quadratic formula: $\operatorname{cosec} \theta = \frac{2\sqrt{3} - 2 \pm \sqrt{(2(\sqrt{3} - 1))^2 + 4 \sqrt{3} \cdot 4}}{2\sqrt{3}}$ Simplifying inside the square root: $(2(\sqrt{3} - 1))^2 + 4\sqrt{3} \cdot 4 = 4(3 + 1 - 2\sqrt{3}) + 16\sqrt{3}$ This simplifies to: $4(4 - 2\sqrt{3}) + 16\sqrt{3} = 16 - 8\sqrt{3} + 16\sqrt{3} = 16 + 8\sqrt{3}$ Therefore, the quadratic gives: $\operatorname{cosec} \theta = \frac{2\sqrt{3} - 2 \pm 2(\sqrt{3} + 1)}{2\sqrt{3}}$ By solving, we find: $\operatorname{cosec} \theta = \frac{-2}{\sqrt{3}}, 2$ Thus, for each value of the cosecant, $\theta$ can take several values that fall within the given interval $[- \frac{7\pi}{6}, \frac{4\pi}{3}]$: $\theta = \frac{-7\pi}{6}, \frac{-2\pi}{3}, \frac{-\pi}{3}, \frac{\pi}{6}, \frac{5\pi}{6}, \frac{4\pi}{3}$ These are all the possible solutions for $\theta$ within the defined range.