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?