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 $\displaystyle \left( \frac{3 \pi}{2} + 2 \pi n, -1 \right)$.