3. Algebraically solve each of the following equations over the domain $0 \leq x \leq 2 \pi$. a) $2 \sin x=-\sqrt{3}$ b) $3 \sec x+6=0$

Step 1 ：Given the equation \(2 \sin x=-\sqrt{3}\), we first isolate \(\sin x\) by dividing both sides by 2 to get \(\sin x = -\frac{\sqrt{3}}{2}\).

Step 2 ：Next, we use the inverse sine function to solve for x. The inverse sine of \(-\frac{\sqrt{3}}{2}\) is \(-60^\circ\) or \(240^\circ\).

Step 3 ：However, the sine function has a period of \(2\pi\), so there may be more than one solution in the given domain.

Step 4 ：The solutions to the equation are \(x = -60^\circ\) and \(x = 240^\circ\). However, the solution \(x = -60^\circ\) is not in the domain \(0 \leq x \leq 2 \pi\). Therefore, the only solution in the given domain is \(x = 240^\circ\).

Step 5 ：Final Answer: The solution to the equation \(2 \sin x=-\sqrt{3}\) over the domain \(0 \leq x \leq 2 \pi\) is \(\boxed{x = 240^\circ}\).