To find the number of solutions to the equation $(4-\sqrt{3}) \sin x - 2 \sqrt{3} \cos^2 x = -\frac{4}{1+\sqrt{3}}, \quad x \in \left[-2\pi, \frac{5\pi}{2}\right]$ we start by letting $\sin x = t$, which implies $\cos^2 x = 1 - t^2$. Substituting these into the equation, we have: $(4-\sqrt{3}) t - 2 \sqrt{3}(1-t^2) = \frac{-4}{1+\sqrt{3}}$ Simplifying, we get: $2 \sqrt{3} t^2 + (4-\sqrt{3}) t - 2\sqrt{3} + \frac{4}{1+\sqrt{3}} = 0$ Further simplification yields: $2 \sqrt{3} t^2 + (4-\sqrt{3}) t + \frac{-2 \sqrt{3} - 2}{1+\sqrt{3}} = 0$ This simplifies to: $2 \sqrt{3} t^2 + (4-\sqrt{3}) t - 2 = 0$ To find $t$, solve the quadratic equation: $t = \frac{(\sqrt{3}-4) \pm \sqrt{19 - 8 \sqrt{3} + 8(2\sqrt{3})}}{4 \sqrt{3}}$ This simplifies to: $t = \frac{(\sqrt{3}-4) \pm \sqrt{19 + 8 \sqrt{3}}}{4 \sqrt{3}} = \frac{(\sqrt{3}-4) \pm (\sqrt{3}+4)}{4 \sqrt{3}}$ Evaluating further, we find: $t = \frac{2 \sqrt{3}}{4 \sqrt{3}} \quad \text{or} \quad \frac{-8}{4 \sqrt{3}}$ This gives us $\sin x = \frac{1}{2}$ or $\frac{-2}{\sqrt{3}} < -1$. Since $\frac{-2}{\sqrt{3}}$ is less than -1, it is not valid for $\sin x$. Hence, the only solution is $\sin x = \frac{1}{2}$. Considering the interval $x \in \left[-2\pi, \frac{5\pi}{2}\right]$, we find that $\sin x = \frac{1}{2}$ occurs at several points. After checking, there are 5 such solutions in the given interval.