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?
 A: Well, since $\lim_{x \rightarrow 0} \frac{sin(x)}{x} = 1$ for every $\epsilon > 0$ you can find an $x_{0}$ such that $sin(x_{0})>x_{0} - \epsilon $. Now use the fact that $b$ is strictly greater than zero. I think this should be enough for you to prove the statement.
If you need more details I can edit the answer, good luck!
A: $f(0)=0$ 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.
A: Using Intermediate Value Theorem: Trivial Solution
We can try solving the question using the Intermediate Value Theorem (here, I obtain the trivial solution $x=0$). Define:
$$
f(x) = x-b\sin{x}
$$
Note that $f(x)$ is continuous for $x \in \mathbb{R}$.
$$
\begin{align}
f(b^2) &= b^2 -b\sin{b^2}>0\\
f(-b^2) &= -b^2 -b\sin{-b^2}\\
&= -b^2 +b\sin{b^2} < 0
\end{align}
$$
Since $f(x)$ is negative at $-b^2$, postive at $b^2$, and continuous in $[-b^2,b^2]$, by Intermediate Value Theorem:
$$
\exists x \in  [-b^2,b^2] \text{ such that } f(x) = 0\\
\Rightarrow \exists x \in [-b^2,b^2] \text{ such that }x=b\sin{x} 
$$
However, as correctly pointed out in the comments, since this $x$ can be zero, it doesn't answer the question exactly. To find a value, not zero, we have to use The Intermediate Value Theorem on two positive values (or two negative values) of $x$.
Using Intermediate Value Theorem: General Solution
We can calculate the value of a $f(x)$ at a value $x=\epsilon>0$ in the $\lim \epsilon \to 0$.
$$
\begin{align}
f(\epsilon) &= \epsilon - b\sin{\epsilon}\\
\Rightarrow \lim_{\epsilon \to 0} f(\epsilon) &= \epsilon - b\epsilon\\
&=\epsilon(1-b) < 0\text{ since }\epsilon>0\text{ and } b >1
\end{align} 
$$
We used the identity $\lim_{\epsilon \to 0}\sin{\epsilon} = \epsilon$.
Thus, the value of $f(x)$ is always negative for values of $x$ close to (but greater than zero). We can now use the Intermediate value theorem to show that $x=b\sin{x}$ for some $x>0$.
