The problem asks to find the two values of $x$ in the interval $[0, 2\pi]$ at which the graph of $f(x) = x - 2 \sin x$ has a horizontal tangent.

Solution Outline:

A horizontal tangent occurs where the derivative of the function $f(x)$ is equal to 0. So, we will:

1. Differentiate $f(x)$.
2. Set the derivative equal to 0 and solve for $x$ within the given interval.

Step 1: Differentiate $f(x)$

We need to compute $f'(x)$, the derivative of $f(x) = x - 2 \sin x$. $f'(x) = \frac{d}{dx}(x) - 2 \cdot \frac{d}{dx}(\sin x) = 1 - 2 \cos x$

Step 2: Set the derivative equal to 0

To find the points where the graph has a horizontal tangent, we set $f'(x) = 0$: $1 - 2 \cos x = 0$ Solve for $\cos x$: $2 \cos x = 1 \quad \Rightarrow \quad \cos x = \frac{1}{2}$

Step 3: Solve for $x$

The general solutions for $\cos x = \frac{1}{2}$ occur at: $x = \frac{\pi}{3} \quad \text{and} \quad x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$

Step 4: Check the interval

Both $x = \frac{\pi}{3}$ and $x = \frac{5\pi}{3}$ lie within the interval $[0, 2\pi]$.

Final Answer:

The two values of $x$ where the graph of $f(x) = x - 2 \sin x$ has a horizontal tangent are: $x = \frac{\pi}{3} \quad \text{and} \quad x = \frac{5\pi}{3}$

Would you like more details or have any questions?

Here are five related questions:

1. What does it mean for a function's graph to have a horizontal tangent?
2. How do we determine the points where a graph has a horizontal tangent for any given function?
3. What other types of tangents can we find apart from horizontal ones?
4. Can you explain how to solve trigonometric equations like $\cos x = \frac{1}{2}$?
5. What role does differentiation play in identifying tangents on a curve?

Tip: When differentiating trigonometric functions, remember that the derivative of $\sin x$ is $\cos x$, and the derivative of $\cos x$ is $-\sin x$.