1. **State the problem:** Find all values of $x$ in the interval $0 < x < 2\pi$ for the function $f(x) = x - 2 \cos x$ where the slope of the tangent line is 2. 2. **Recall the formula:** The slope of the tangent line to the graph of $f(x)$ at any point $x$ is given by the derivative $f'(x)$. 3. **Find the derivative:** $$f'(x) = \frac{d}{dx}\left(x - 2 \cos x\right) = 1 - 2(-\sin x) = 1 + 2 \sin x$$ 4. **Set the derivative equal to the slope:** We want $f'(x) = 2$, so $$1 + 2 \sin x = 2$$ 5. **Solve for $\sin x$:** $$2 \sin x = 2 - 1 = 1$$ $$\sin x = \frac{1}{2}$$ 6. **Find $x$ values where $\sin x = \frac{1}{2}$ in $0 < x < 2\pi$:** The solutions are $$x = \frac{\pi}{6}, \frac{5\pi}{6}$$ 7. **Conclusion:** The values of $x$ where the tangent line to $f(x)$ has slope 2 in the interval $0 < x < 2\pi$ are $$x = \frac{\pi}{6}, \frac{5\pi}{6}$$