Intermediate value theorem to show there are at least two values of x

Question

I have a really straightforward equation which says:

''Show that there are at least two values of $x$ in the interval $(-\pi,\pi)$ such that $x=1+\sin(x)$''

Easy enough. So I defined $h(x)=x-1-\sin(x)$applied the intermediate value theorem because if I can show $h(x)=0$ then that would imply $x=1+\sin(x)$.

So I plugged in $x=0$, I get $h(0)=-1$ and if I plug in $x=\pi$ , I get $h(\pi)=\pi -1$ which is positive so that would imply there is a root for $h(x)$ and therefore, $x=1+\sin(x)$.

However, I can't show there is another x value that satisfies this. Wolfram alpha only shows there is 1 point of intersection so I don't know how the question can show that there is a second intersection point for $x=1+\sin(x)$ because there is only ever 1 intersection point so I am a little confused...

Is there a second point of intersection? Or is the question worded incorrectly?

Answer 1

Indeed, there is only one solution to $x=1+\sin(x)$ in $(-\pi,\pi)$. Their graphs look like this: