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 \(\boxed{\left\{\frac{2 \pi}{3}, \frac{5 \pi}{6}\right\}}\).