Show that $\forall b > 1$, $\exists x \neq 0$ such that $x = b \sin x$ This is a problem from my calculus / analysis course.

Show that $\forall b > 1$, $\exists x \neq 0$ such that $x = b \sin x$.

My first thought is to use the intermediate value theorem:
Pick any $b>1$.
Let $f(x) = x - b \sin x$. Clearly $f$ is continuous on $\mathbb{R}$.
Then $f(b^2) = b^2 - b \sin (b^2) > 0$.
Now, I need to find some $c > 0$ such that $f(c) < 0$.
Then by the intermediate value theorem, I can show that there exists some $x \in [c,b^2]$ such that $f(x) = 0$, which is the desired result.
However, I cannot find such $c$. Is there any hint on this? Or are there any other approaches to this question?

$f(0)=0$ and $f'(0)=1-b<0$. By the definition of the derivative, there exists $x>0$ sufficiently small such that
$$ \frac{1-b}2<\frac{f(x)-f(0)}x -(1-b)<\frac{-(1-b)}2 $$
The right inequality can be rewritten $$ \frac{f(x)}x< \frac{1-b}2$$
i.e. $f(x) < \frac{(1-b)x}2 < 0$. Now you can finish with IVT.
