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We are tasked with finding the \( x \)-values where the tangent line to the function \( f(x) = x - 2\cos(x) \) has a slope of 1.

### Step-by-Step Solution:

1. **Find the derivative of the function \( f(x) \):**
   The slope of the tangent line is given by the derivative \( f'(x) \).

   \[
   f'(x) = \frac{d}{dx} \left( x - 2\cos(x) \right)
   \]
   The derivative of \( x \) is 1, and the derivative of \( -2\cos(x) \) is \( 2\sin(x) \), so:

   \[
   f'(x) = 1 + 2\sin(x)
   \]

2. **Set the derivative equal to 1:**
   Since we are looking for the points where the slope of the tangent line is 1, set \( f'(x) = 1 \):

   \[
   1 + 2\sin(x) = 1
   \]

3. **Solve for \( \sin(x) \):**
   Simplifying the equation:

   \[
   2\sin(x) = 0
   \]

   \[
   \sin(x) = 0
   \]

4. **Find the \( x \)-values where \( \sin(x) = 0 \):**
   The sine function equals 0 at multiples of \( \pi \):

   \[
   x = n\pi \quad \text{for} \quad n \in \mathbb{Z}
   \]

### Conclusion:
The \( x \)-values where the tangent line to the graph of \( f(x) = x - 2\cos(x) \) has a slope of 1 are \( x = n\pi \), where \( n \) is any integer.

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Would you like further details on any step? Here are some related questions to consider:

1. How do you find the derivative of trigonometric functions?
2. Why does the sine function equal 0 at multiples of \( \pi \)?
3. What does the slope of the tangent line tell us about the behavior of the function at that point?
4. Can the tangent slope be negative for this function? Why or why not?
5. How do you confirm that these \( x \)-values correspond to points on the graph of the original function?

**Tip:** To solve equations involving trigonometric functions, it's useful to recall the basic properties and periodic nature of these functions.

Find all x values on the graph of f(x) = x - 2cos(x) where the tangent line has slope 1.

Learn how to find all x-values where the tangent line to the graph of f(x) = x - 2cos(x) has a slope of 1. This problem involves differentiation and basic trigonometric properties. Suitable for students in Grades 11-12.

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