1. **Problem statement:** Find the number of points on the interval $[-5,5]$ where the tangent line to the curve $y = x + \cos x$ is parallel to the secant line over the interval $[-5,5]$. 2. **Step 1: Find the slope of the secant line.** The secant line connects the points at $x = -5$ and $x = 5$ on the curve. Calculate $y(-5)$ and $y(5)$: $$y(-5) = -5 + \cos(-5) = -5 + \cos 5$$ $$y(5) = 5 + \cos 5$$ The slope of the secant line is: $$m_{sec} = \frac{y(5) - y(-5)}{5 - (-5)} = \frac{(5 + \cos 5) - (-5 + \cos 5)}{10} = \frac{5 + \cos 5 + 5 - \cos 5}{10} = \frac{10}{10} = 1$$ 3. **Step 2: Find the slope of the tangent line.** The slope of the tangent line at any point $x$ is the derivative of $y$: $$y' = \frac{d}{dx}(x + \cos x) = 1 - \sin x$$ 4. **Step 3: Set the tangent slope equal to the secant slope to find points where they are parallel:** $$1 - \sin x = 1$$ $$-\sin x = 0$$ $$\sin x = 0$$ 5. **Step 4: Solve $\sin x = 0$ on the interval $[-5,5]$.** The solutions to $\sin x = 0$ are at $x = n\pi$ where $n$ is an integer. Find all integers $n$ such that $n\pi \in [-5,5]$: Since $\pi \approx 3.14159$, possible $n$ values are $-1, 0, 1$ because: $$-5 \leq n\pi \leq 5 \Rightarrow -5/\pi \leq n \leq 5/\pi$$ $$-1.59 \leq n \leq 1.59$$ So $n = -1, 0, 1$. Corresponding $x$ values: $$x = -\pi, 0, \pi$$ 6. **Step 5: Count the number of points:** There are 3 points where the tangent is parallel to the secant line. **Final answer:** (D) 3