Suppose $f$ is odd and periodic. Show that the graph of $f$ crosses the $x-$axis infinitely often. 
Suppose $f$ is odd and periodic. Show that the graph of $f$ crosses the $x-$axis infinitely often.

I don't understand the proof of this 
 solution is :
$f$ is odd $\implies f(0)=−f(0)\implies f(0)=0.$
So $f(c)=f(2c)=\cdots=0,$ also (by periodicity, where $c$ is the period)
 A: For any odd function, $f(-x) = -f(x)$ for all $x$. Plugging in $x = 0$ gives us $$f(0) = f(-0) = -f(0)$$so we know $f(0) = 0$. Now we know that our function is periodic with period $c$. Then $f(x) = f(x+c)$ for all $x$. From these two observations, we note that $$f(c) = f(0 + c) = f(0) = 0$$and likewise,
$$f(2c) = f(c + c) = f(c) = 0$$ and so on.
A: A function $f: \mathbb R \to \mathbb R$ is said to be odd if $f(x)=-f(-x)$. 
In particular, for $x=0$, this implies that $f(0)=-f(0)$. The only time that a number is the same as its negative in $\mathbb R$ is if it is itself zero. Hence, $f(0)=0$. Then being periodic implies that $f(x)=f(x+c)$ for some $c \in \mathbb R$ and for all $x \in \mathbb R$. Hence, $0=f(0)=f(c)=f(2c)=\dots$
A: As $f$ is an odd function, therefore by definition $$f(-x)=-f(x)$$ for all $x.$ 
In particular, taking $x=0,$ we get $$f(-0)=-f(0)\\f(0)=0$$
Let the period of $f$ be $c$.
Then $$f(x+c)=f(x)$$ for all $x.$ In particular, $$f(0+c)=f(0)\\f(c)=0$$
Also, $$f(c+c)=f(c)\\f(2c)=0$$
Proceeding this way $$f(nc)=0$$ for all $n \in \mathbb Z.$
Thus, $f$ cuts the $x-$axis infinitely often.
A: That's not true. Let us take the function
$$
f(x)=\frac{\sin^2x}{\sin x}\,,\qquad x\neq k\pi,\ k\in\mathbb Z.
$$
It is an odd function, however it does not cross the $x$-axis at any point.
