The solution set for the equation $\sin x=-\frac{\sqrt{3}}{2}$ is $\left\{\frac{4 \pi}{3}, \frac{5 \pi}{3}\right\}$. Use this information to solve the following equation over the interval $[0,2 \pi)$.
\[
\sin 2 x=-\frac{\sqrt{3}}{2}
\]
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. The solution set is \{\}
(Simplify your answer. Type an exact answer, using $\pi$ as needed. Use integers or fractions for any numbers in the expression. Use a comma to
B. The solution set is $\varnothing$.

Step 1 ：The given equation is \(\sin 2x = -\frac{\sqrt{3}}{2}\). We know that the solution set for the equation \(\sin x = -\frac{\sqrt{3}}{2}\) is \(\left\{\frac{4 \pi}{3}, \frac{5 \pi}{3}\right\}\). Therefore, we can set \(2x = \frac{4 \pi}{3}\) and \(2x = \frac{5 \pi}{3}\) to find the solutions for x.

Step 2 ：Solving these equations gives us the solutions \(x = \frac{2 \pi}{3}\) and \(x = \frac{5 \pi}{6}\).

Step 3 ：We need to check if these solutions are within the interval \([0,2 \pi)\).

Step 4 ：Both solutions, \(x = \frac{2 \pi}{3}\) and \(x = \frac{5 \pi}{6}\), are within the interval \([0,2 \pi)\).

Step 5 ：Therefore, the solution set for the equation \(\sin 2x = -\frac{\sqrt{3}}{2}\) over the interval \([0,2 \pi)\) is \(\left\{\frac{2 \pi}{3}, \frac{5 \pi}{6}\right\}\).

Step 6 ：So, the final answer is \(\boxed{\left\{\frac{2 \pi}{3}, \frac{5 \pi}{6}\right\}}\).