To find all $x$ values where the tangent line to the graph of the function $f(x) = x - 2\cos(x)$ has a slope of 1, we follow these steps:

1. Take the derivative of $f(x)$

The derivative of a function gives the slope of the tangent line at any point $x$. The derivative of $f(x)$ is:

$f'(x) = \frac{d}{dx} \left( x - 2\cos(x) \right)$

Using basic differentiation rules:

$f'(x) = 1 + 2\sin(x)$

2. Set the derivative equal to 1

Since the slope of the tangent line is given as 1, we set $f'(x) = 1$:

$1 + 2\sin(x) = 1$

3. Solve for $x$

Simplify the equation:

$2\sin(x) = 0$

$\sin(x) = 0$

Now, solve for $x$. The general solutions for $\sin(x) = 0$ are:

$x = n\pi \quad \text{where} \quad n \in \mathbb{Z}$

4. Conclusion

The values of $x$ where the tangent line to the graph of $f(x) = x - 2\cos(x)$ has a slope of 1 are:

$x = n\pi \quad \text{where} \quad n \in \mathbb{Z} \ (\text{any integer}).$

Would you like more details or have any questions on this?

Related questions:

1. How would the solution change if the slope of the tangent line were different?
2. What is the significance of $n\pi$ in terms of periodicity on the graph?
3. How do we determine the concavity of $f(x) = x - 2\cos(x)$?
4. What does the second derivative tell us about the behavior of $f(x)$?
5. How can you graphically interpret the result $x = n\pi$?

Tip:

The sine function $\sin(x) = 0$ repeats periodically, which is why the solution is valid for any integer $n$. Keep this in mind when solving trigonometric equations.