Solve the equation on the interval $[0,2 \pi)$. \[ \sin 2 x=-\sqrt{3} \sin x \] Select the correct choice below and, if necessary, fill in the answer box to complete your choice A. The solution set is (Type your answer in radians. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.) B. The solution is the empty set.

Step 1 ：By the double-angle formula, \(\sin 2x = 2 \sin x \cos x\), so \(\sin 2x = -\sqrt{3} \sin x\) becomes \(2 \sin x \cos x = -\sqrt{3} \sin x\).

Step 2 ：Moving everything to one side, and taking out a factor of \(\sin x\), we get \(\sin x (2 \cos x + \sqrt{3}) = 0\).

Step 3 ：We have that \(\sin x = 0\) for \(x = 0\), \(\pi\), and \(2 \pi\), and \(\cos x = -\frac{\sqrt{3}}{2}\) for \(x = \frac{4 \pi}{3}\) and \(x = \frac{5 \pi}{3}\).

Step 4 ：Thus, the solution set is \(\boxed{\{0, \pi, 2\pi, \frac{4\pi}{3}, \frac{5\pi}{3}\}}\).