Find the horizontal tangent line Fine the $x$-coordinate of all points on the curve $y=\sin(2x)+2\sin(x)$ at which the tangent line is horizontal. Consider the domain $x=[0,2\pi)$. There are three $x$ values.
Can you explain how to find the horizontal tangent line and vertical tangent line.
Also, $x=[0,2\pi)$ means that $0$ is not included?

The gradient$(m)$ of the tangent line $=f'(x)$ 
The tangent line will be horizontal of $y=f(x)$ if $f'(x)=0$ and will be vertical if $\displaystyle f'(x)=\infty\implies \frac1{f'(x)}=0$ 
Now, here $\displaystyle y=\sin2x+2\sin x\implies \frac{dy}{dx}=2\cos2x+2\cos x$
So, horizontal tangent, we need $\cos2x+\cos x=0\iff \cos2x=-\cos x=\cos(\pi-x)$ 
$\displaystyle\implies 2x=2n\pi\pm(\pi-x)$ where $n$ is any integer
Taking the '+' sign, $\displaystyle\implies 2x=2n\pi+(\pi-x)\implies x=\frac{(2n+1)\pi}3$
Now $0\le x<2\pi\implies 0\le \frac{(2n+1)\pi}3<2\pi\iff 0\le 2n+1<6\iff0\le n\le2$
Similarly, for the '-' sign
Now as $\displaystyle|\cos y|\le1$ for real $\displaystyle y, \frac{dy}{dx}=2\cos2x+2\cos x$ will be finite, hence no vertical tangent line
In fact, we can find the range of $\displaystyle 2\cos2x+2\cos x$ for real $x$
