Find maxima and minima of $f(x) = x \cot x$ So I got a function

$$f(x)= x \cot x $$

I would like to find values of $x$ where $f'(x) = 0$
Applying product rule, we get:
$$f'(x) = \cot x - x \cdot csc^2 x $$
Setting equation to zero
$$\cot x - x \cdot csc^2 x = 0 $$
Now, I will try my best at simplifying equation above:
$$\frac{\cos x}{\sin x} - x \cdot \frac{1}{\sin^2x} = 0$$
Provided that $x ≠ 0$, multiply $\frac{cos x}{\sin x}$ by $\frac{\sin x}{\sin x}$
$$\frac{\cos x \sin x}{\sin^2 x} - \frac{x}{\sin^2x} = 0 \implies$$ 
$$\frac{\cos x \sin x - x }{\sin^2 x}= 0 $$
Multiplying both sides by $\sin^2 x$, we arrive at
$$\cos x \sin x - x = 0 $$
Now, assuming that all calculations above are correct, we can make some observations:
If $x ≥ 1 $ or $x ≤ -1$, then $\cos x \sin x + x ≠ 0 $. Thus if equation above has roots, then $x$ must be somewhere between $(-1,1)$
Equation above equals zero if $x = 0$, but we stipulated before that $x$ cannot equal $0$.
From here, I am stuck.
By looking at the graphing calculator, it's getting evident that there is 
no value $x$ such that $f'(x) = 0$, but how do I show it mathematically?

Consider $g(x) = \cos x \sin x -x$, then $g(0)=0$ and $g(x)$ is monotone because $g'(x) = -2\sin^2(x)$.
