Single Variable Calculus, Chapter 3, 3.4, Section 3.4, Problem 33

Find what values of $x$ does the graph of $f(x) = x + 2 \sin x$ have a horizontal tangent.

Solving for $f'(x)$


$\begin{equation} \begin{aligned} f'(x) =& \frac{d}{dx} (x) + 2 \frac{d}{dx} (\sin x) && \text{} \\ \\ f'(x) =& 1 + 2 \cos x && \text{} \\ \\ m_T =& 0 \qquad \text{ slope of the tangent is horizontal} && \\ \\ \end{aligned} \end{equation}$


Let $f'(x) = m_T$ (slope of the tangent line)


$\begin{equation} \begin{aligned} f'(x) = m_T =& 1 + 2 \cos x \\ \\ 0 =& 1 + 2 \cos x \\ \\ 2 \cos x =& -1 \\ \\ 2 \cos x =& \frac{-1}{2} \end{aligned} \end{equation}$


By using the unit circle diagram, we can determine what angle(s) has $\displaystyle \frac{-1}{2}$ on $x$-coordinate, so..


$\begin{equation} \begin{aligned} x =& \cos^{-1} \left[ \frac{-1}{2} \right] \\ \\ x =& \frac{2}{3} \pi \text{ and } x = \frac{4}{3} \pi \end{aligned} \end{equation}$


Also, we know that the trigonometric functions have repeating cycles so the answer is

$\displaystyle x = \frac{4}{3} \pi + 2 \pi (n)$ and $\displaystyle x = \frac{2}{3} \pi + 2 \pi (n)$ ; where $n$ is any integer and $2 \pi$ corresponds to the repeating period.