1. **State the problem:** Find the zeros of the second derivative $g''(x)$ given by $$g''(x) = \frac{-2\sin x + 2\sin x \cos x}{(2+\cos x)^3} = 0$$ 2. **Set numerator equal to zero and denominator not zero:** $$-2\sin x + 2\sin x \cos x = 0 \quad \text{and} \quad (2 + \cos x)^3 \neq 0$$ 3. **Factor the numerator:** $$2 \sin x (-1 + \cos x) = 0$$ 4. **Solve each factor equal to zero:** - $2 \sin x = 0 \implies \sin x = 0$ - $-1 + \cos x = 0 \implies \cos x = 1$ 5. **Check denominator condition:** $2 + \cos x \neq 0 \implies \cos x \neq -2$ (always true since $\cos x \in [-1,1]$) 6. **Find solutions:** - $\sin x = 0 \implies x = k\pi, k \in \mathbb{Z}$ - $\cos x = 1 \implies x = 2k\pi, k \in \mathbb{Z}$ Since $\cos x = 1$ is a subset of $\sin x = 0$ solutions, the zeros of $g''(x)$ are: $$x = k\pi, \quad k \in \mathbb{Z}$$ **Final answer:** $$\boxed{x = k\pi, \quad k \in \mathbb{Z}}$$