1. **State the problem:** Solve the trigonometric equation $$\cos^2 x + 2 \sin x + \sin^2 x = 0$$. 2. **Use the Pythagorean identity:** Recall that $$\cos^2 x + \sin^2 x = 1$$. 3. **Substitute the identity into the equation:** $$\cos^2 x + \sin^2 x + 2 \sin x = 0 \implies 1 + 2 \sin x = 0$$ 4. **Isolate $$\sin x$$:** $$2 \sin x = -1 \implies \sin x = -\frac{1}{2}$$ 5. **Find the general solutions for $$\sin x = -\frac{1}{2}$$:** The sine function equals $$-\frac{1}{2}$$ at angles where $$x = 7\pi/6 + 2k\pi \quad \text{or} \quad x = 11\pi/6 + 2k\pi$$ for any integer $$k$$. 6. **Final answer:** $$x = \frac{7\pi}{6} + 2k\pi \quad \text{or} \quad x = \frac{11\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}$$