1. **State the problem:** Solve the equation $$\csc x + 2 = 0$$ for $$0 \leq x < 2\pi$$. 2. **Rewrite the equation:** Recall that $$\csc x = \frac{1}{\sin x}$$, so the equation becomes: $$\frac{1}{\sin x} + 2 = 0$$ 3. **Isolate $$\sin x$$:** $$\frac{1}{\sin x} = -2$$ 4. **Invert both sides:** $$\sin x = \frac{1}{-2} = -\frac{1}{2}$$ 5. **Find all $$x$$ in $$[0, 2\pi)$$ where $$\sin x = -\frac{1}{2}$$:** - The sine function equals $$-\frac{1}{2}$$ at angles in the third and fourth quadrants. - These angles are: $$x = \frac{7\pi}{6}, \frac{11\pi}{6}$$ 6. **Final answer:** $$x = \frac{7\pi}{6}, \frac{11\pi}{6}$$ These correspond to option b.