1. The problem is to find the x-coordinates of the inflection points of the function $f(x) = \sin(x)$ on the interval $[0, 2\pi]$. 2. Inflection points occur where the second derivative changes sign, i.e., where $f''(x) = 0$ and the concavity changes. 3. Compute the first derivative: $$f'(x) = \cos(x)$$ 4. Compute the second derivative: $$f''(x) = -\sin(x)$$ 5. Set the second derivative equal to zero to find potential inflection points: $$-\sin(x) = 0 \Rightarrow \sin(x) = 0$$ 6. On the interval $[0, 2\pi]$, $\sin(x) = 0$ at $x = 0, \pi, 2\pi$. 7. Check concavity change by examining the sign of $f''(x)$ around these points: - For $x$ just less than $\pi$, $\sin(x)$ is positive, so $f''(x) = -\sin(x)$ is negative (concave down). - For $x$ just greater than $\pi$, $\sin(x)$ is negative, so $f''(x) = -\sin(x)$ is positive (concave up). Thus, concavity changes at $x = \pi$, indicating an inflection point. At $x=0$ and $x=2\pi$, the concavity does not change sign, so these are not inflection points. 8. Therefore, the only inflection point on $[0, 2\pi]$ is at $x = \pi$. Final answer: C) $x = \pi$