Single Variable Calculus, Chapter 3, 3.4, Section 3.4, Problem 34

Determine the points on the curve $\displaystyle y = \frac{\cos x}{2 + \sin x}$ at which the tangent is horizontal.


$
\begin{equation}
\begin{aligned}

y' =& \frac{\displaystyle (2 + \sin x) \frac{d}{dx} (\cos x) - \left[ (\cos x) \frac{d}{dx} (2 + \sin x) \right] }{(2 + \sin x)^2}
&& \text{Apply Quotient Rule}
\\
\\
y' =& \frac{(2 + \sin x)(- \sin x) - (\cos x) (0 + \cos x)}{(2 + \sin x)^2}
&& \text{Expand the equation}
\\
\\
y' =& \frac{-2 \sin x - \sin ^2 x - \cos^2 x}{4 + 2 \sin x + \sin^2 x}
&& \text{Group the $(- \sin^2 x - \cos^2 x)$ terms together and factor out the negative sign.}
\\
\\
y' =& \frac{-2 \sin x - (\sin ^2 x + \cos^2 x)}{4 + 2 \sin x + \sin^2 x}
&& \text{Apply Pythagorean Identity, $\sin ^2 x + \cos^2 x = 1$}
\\
\\
y' =& \frac{-2 \sin x - 1}{4 + 2 \sin x + \sin^2 x}
&& \text{Since the slope is horizontal so the slope of the tangent line is zero. Let $y' = m_T$(slope of the tangent line)}
\\
\\
0 =& \frac{-2 \sin x - 1}{4 + 2 \sin x + \sin ^2 x}
&& \text{Multiply both sides by $(4 + 2 \sin x + \sin^2 x)$}
\\
\\
-2 \sin x - 1 =& 0
&& \text{Add 1 to both sides}
\\
\\
-2 \sin x =& 1
&& \text{Divide both sides by -2}
\\
\\
\sin x =& \frac{-1}{2}


\end{aligned}
\end{equation}
$



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


$
\begin{equation}
\begin{aligned}

x =& \sin^- \left[ \frac{-1}{2} \right]
\\
\\
x =& \frac{7 \pi}{6} \text{ and } x = \frac{11 \pi}{6}

\end{aligned}
\end{equation}
$


Also, recall that sine function has repeating cycles every period of $2 \pi$, so the answer is

$\displaystyle x = \frac{7 \pi}{6} + 2 \pi (n)$ and $\displaystyle x = \frac{11 \pi}{6} + 2 \pi (n)$ where $n$ is any integer and $2 \pi$ corresponds to the repeating period.