1. **State the problem:** Verify Rolle's theorem for the function $f(x) = \sin^2(x)$ on the interval $[0, \pi]$. 2. **Recall Rolle's theorem:** If a function $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a) = f(b)$, then there exists at least one $c \in (a,b)$ such that $f'(c) = 0$. 3. **Check conditions:** - $f(x) = \sin^2(x)$ is continuous and differentiable everywhere. - Evaluate endpoints: $f(0) = \sin^2(0) = 0$, $f(\pi) = \sin^2(\pi) = 0$. - Since $f(0) = f(\pi)$, conditions are met. 4. **Find derivative:** $$f'(x) = 2\sin(x)\cos(x) = \sin(2x)$$ 5. **Find $c$ such that $f'(c) = 0$:** $$\sin(2c) = 0 \implies 2c = n\pi, \quad n \in \mathbb{Z}$$ 6. **Find $c$ in $(0, \pi)$:** $$c = \frac{n\pi}{2}$$ For $n=1$, $c = \frac{\pi}{2} \in (0, \pi)$. 7. **Conclusion:** At $c = \frac{\pi}{2}$, $f'(c) = 0$, so Rolle's theorem is verified for $f(x) = \sin^2(x)$ on $[0, \pi]$.