Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 60

Determine the $x$-coordinates of all points on the curve $y = \sin 2 x - 2 \sin x$ at which the tangent line is horizontal.


$
\begin{equation}
\begin{aligned}

y' = m =& \frac{d}{dx} (\sin 2 x - 2 \sin)
&& \text{Where $m = 0$ because tangent line is horizontal}
\\
\\
m =& \frac{d}{dx} (\sin 2 x) - 2 \frac{d}{dx} \sin
&&
\\
\\
m =& \cos 2 x \cdot \frac{d}{dx} (2x) - 2 \cos x
&&
\\
\\
m =& (\cos 2x) (2) - 2 \cos x
&&
\\
\\
0 =& 2 \cos 2x - 2 \cos x
&&
\\
\\
2 \cos 2x =& 2 \cos x
&&
\\
\\
\frac{\cancel{2} \cos 2x}{\cancel{2}} =& \frac{\cancel{2} \cos x}{\cancel{2}}
&&

\end{aligned}
\end{equation}
$




$
\begin{equation}
\begin{aligned}

& \cos 2 x = \cos x
&& \text{Using Double Angle Formula $(\cos 2x = 2 \cos^2 x - 1)$}
\\
\\
& 2 \cos^2 x - 1 = \cos x
&&
\\
\\
& 2 \cos ^2 x - \cos x - 1 = 0
&&
\\
\\
& 2 \cos ^2 x - 2 \cos x + \cos x - 1 = 0
&& \text{We let $- \cos x = -2 \cos x + \cos x$ to have a complete factor}
\\
\\
& 2 \cos x (\cos x - 1) + 1 (\cos x - 1) = 0
&&
\\
\\
& (2 \cos x + 1) (\cos x - 1) = 0
&&

\end{aligned}
\end{equation}
$


$\displaystyle \cos x = \frac{-1}{2} || \cos x = 1$







Based from the graph and unit circle diagram

$\displaystyle x = 2 \pi n, \frac{\pi n}{2} \pm \frac{\pi}{3} \qquad $ (where $n$ is any integer)