The graph of $f(x) = x + 2 \sin(x)$ will have a horizontal tangent where the derivative $f'(x)$ is equal to zero, as a horizontal tangent indicates a slope of zero.

1. Find the derivative $f'(x)$:

   $f(x) = x + 2 \sin(x)$ Differentiating with respect to $x$, we get: $f'(x) = 1 + 2 \cos(x)$

2. Set the derivative equal to zero to find the horizontal tangents:

   $1 + 2 \cos(x) = 0$

3. Solve for $\cos(x)$:

   $2 \cos(x) = -1$ $\cos(x) = -\frac{1}{2}$

4. Determine the values of $x$ that satisfy $\cos(x) = -\frac{1}{2}$:

   The cosine function equals $-\frac{1}{2}$ at angles: $x = \frac{2\pi}{3} + 2n\pi \quad \text{and} \quad x = \frac{4\pi}{3} + 2n\pi$ where $n$ is any integer.

Final Answer:

The values of $x$ for which the graph of $f(x) = x + 2 \sin(x)$ has a horizontal tangent are: $x = \frac{2\pi}{3} + 2n\pi \quad \text{and} \quad x = \frac{4\pi}{3} + 2n\pi \quad \text{for integers } n.$

Would you like a detailed step-by-step explanation, or do you have any questions?

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1. How can we confirm that these points correspond to horizontal tangents on the original function?
2. What are the implications of finding where $f'(x) = 0$ in terms of identifying maxima or minima?
3. How would the behavior of $f(x) = x + 2 \sin(x)$ change if we altered the coefficient of $\sin(x)$?
4. What role does the period of the cosine function play in finding all solutions?
5. Could there be any points where the tangent is undefined, and how would we identify them?

Tip: For periodic functions like sine and cosine, identifying general solutions often involves adding $2n\pi$ to account for all occurrences across intervals.