Show that $\frac{\mathrm{d} }{\mathrm{d} x}(\sin^n x \sin(nx))= n \sin ^{n-1} x \sin((n+1)x))$ 
Assume that $n\geq 2 $. Show that
$$\frac{\mathrm{d} }{\mathrm{d} x}(\sin^n x \sin(nx))= n \sin ^{n-1} x \sin((n+1)x))$$

At what points $x$ in $[0,\pi]$ does the graph of $y=\sin^n x \sin(nx)$ have a horizontal tangent? 


Of course, I did my work. But I got a pretty weird answer at the very end, so I am hoping stacks can check my answer. 

$$\frac{\mathrm{d} }{\mathrm{d} x}(\sin^n x \sin(nx))= n \sin ^{n-1} x \sin((n+1)x))$$
  is pretty trivial. I have no problems with it. Consider it shown.


To have a horizontal tangent, the value of $\frac{\mathrm{dy} }{\mathrm{d} x} $ must be $0$ at that point which implies, 
$$n \sin ^{n-1} x \sin((n+1)x)) =0 $$
Since $n\geq 0$, either $\sin ^{n-1}x=0$      or     $\sin((n+1)x)=0$
Solving for $\sin ^{n-1}x=0$, we get $x=0$ or $x=\pi$. 
Solving for $\sin((n+1)x)=0$, we get $$x=0\ or\ x=\frac {a\pi} {n+1},\text{ where }a\in {\mathbb{Z^+}}\text{ such that }a\leq n+1\text{ for } n\geq2$$
In another words, there are multiple points where the graph has a horizontal tangent and the number of points depends on the value of $n$.



I have never experienced such an answer in my homework before, Could it be possibly, an error in my workings?
 A: Your work looks correct. There are indeed multiple points where the function has a horizontal tangent, and those points depend on $n$.
One little thing: If $\sin((n+1)x) = 0$, then you can conclude that $(n+1)x = a\pi$ for any $a\in \mathbb{Z}$. Hence in all you get horizontal tangents at the points $x = \frac{a\pi}{n+1}$ for any $a\in \mathbb{Z}$ and also for $x = k\pi$ for $k\in \mathbb{Z}$. But since you are considering only $x\in [0, \pi]$ you get the solutions that you stated.
Another little thing, you are assuming that $n\geq2$ and not $n\geq 0$ as it is written currently.
Another example where this happens if with $f(x) = \sin(nx)$, again here $f'(x) = n\cos(nx)$ and so if $n\geq 1$, say, you would have a horizontal tangent for $nx$ being a multiple of $\pi$ plus $\frac{\pi}{2}$: $nx = k\pi + \frac{\pi}{2}$.
A: You can make yourself more confident that you have the right answer by plugging in some common numbers for $n$. For example:
For $n=2$, $\sin((2+1)x) = 0$ when $x = \frac{\pi}{3}, \frac{2\pi}{3}, \pi,\dots$
For $n=3$, $\sin((3+1)x) = 0$ when  $x = \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi,\ldots$
And so on.
More visually, think of what the graph of $\sin(x)$ looks like and what the transformation $\sin(n \cdot x)$ does to the graph: It makes the wave have a higher frequency. Within the same interval $[0,\pi]$, the graph of $\sin(n \cdot x)$ is going to have $n$ as many zeroes as $\sin(x)$.
