Prove $f(x)$ has three extremum points on $(-2\pi,2\pi)$ Prove $f(x)$ has three extremum points on $(-2\pi,2\pi)$
$$f(x) = \frac{\sin x}{x}, x \not= 0 $$
$$f(x) = 1, x = 0 $$
I calculated its derivative.
$$f'(x) = \frac{x\cos x-\sin x}{x^2}, x \not= 0$$
$$f'(x) = 0, x = 0$$
I can see that $f'(x)$ has a solution for x = 0.
I am not sure how to solve for $x\not= 0$ though. I have tried this:
$$ \frac{x\cos x - \sin x}{x^2} = 0 $$
Multiply both sides by $x^2$
$$ x\cos x - \sin x = 0 $$
Add $\sin x$ to both sides
$$ x\cos x = \sin x$$
Now divide both sides by $\cos x$
$$ x = \tan x $$
Did I do anything wrong? Can this equation be figured out without using any kind of graphing tool?

What you did is very correct and there is no explicit solution to the equation.
Discarding the trivial case $x=0$, it is better to consider that you look for the zero's of
$$f(x)=x \cos (x)-\sin (x)$$ Now, consider a Taylor expansion close to $\frac {3 \pi} 2$ (remember the symmetry) to get
$$f(x)=1+\frac{3}{2} \pi  \left(x-\frac{3 \pi }{2}\right)+\frac{1}{2} \left(x-\frac{3 \pi
   }{2}\right)^2-\frac{1}{4} \pi  \left(x-\frac{3 \pi }{2}\right)^3+O\left(\left(x-\frac{3 \pi   }{2}\right)^4\right)$$ and use series reversion to get
$$x=\frac{3 \pi }{2}-\frac{2}{3 \pi }-\frac{16}{81 \pi ^3}-\frac{16}{243 \pi ^5}- \cdots$$ This truncated expansion would give $x=4.4936$ while the "exact" solution would be $4.4934$.
Just compute the value of the function at this point.
