Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 59

Determine all the points on the graph of the function $f(x) = 2 \sin x + \sin ^2 x $ at which the tangent line is horizontal.


$
\begin{equation}
\begin{aligned}

f'(x) = m =& \frac{d}{dx} (2 \sin x + \sin ^2 x )
\qquad \qquad \text{Where $m = 0$ because tangent line is horizontal}
\\
\\
m =& 2 \frac{d}{dx} (\sin x) + \frac{d}{dx} (\sin x)^2
\\
\\
m =& 2 \cos x + 2 \sin x \cdot \frac{d}{dx} (\sin x)
\\
\\
m =& 2 \cos x (1 + \sin x)
\\
\\
0 =& 2 \cos x (1 + \sin x)



\end{aligned}
\end{equation}
$


$
\begin{array}{c||c}
2 \cos x = 0 & \sin x + 1 = 0 \\
\displaystyle \frac{2 \cos x}{2} = \frac{0}{2} & \sin x = -1 \\
\cos x = 0 &
\end{array}
$

Using the Unit Circle Diagram







$\displaystyle x = \frac{\pi}{2} + 2 \pi n, \qquad \qquad x = \frac{3 \pi}{2} + 2 \pi n \qquad \qquad$ (where $2 \pi n$ refers to the succeeding periods where $n$ is any number)

Solving for $y$


$
\begin{equation}
\begin{aligned}

f \left( \frac{\pi}{2} \right) =& 2 \sin x + \sin^2 x
\\
\\
f \left( \frac{\pi}{2} \right) =& 2 \sin \left( \frac{\pi}{2} \right) + \sin ^2 \left( \frac{\pi}{2} \right)
\\
\\
f \left( \frac{\pi}{2} \right) =& 2 + 1
\\
\\
f \left( \frac{\pi}{2} \right) =& 3
\\
\\
f \left( \frac{3 \pi}{2} \right) =& 2 \sin x + \sin ^2 (x)
\\
\\
f \left( \frac{3 \pi}{2} \right) =& 2 \sin \left( \frac{3 \pi}{2} \right) + \sin ^2 \left( \frac{3 \pi}{2} \right)
\\
\\
f \left( \frac{3 \pi}{2} \right) =& -2 + 1
\\
\\
f \left( \frac{3 \pi}{2} \right) =& -1

\end{aligned}
\end{equation}
$


The points on the given function which the tangent is horizontal are $\displaystyle \left( \frac{3 \pi}{2} + 2 \pi n, -1 \right) $.