What are the solutions of $x=\cot x$? Need to find intervals in which the function $y=\frac{x}{2}\cdot \cos x$ is increasing and decreasing. I tried to solve it on the way below but don't know how to continue.
$ \\ 
y=\frac{x}{2}\cdot \cos x,\ x\in (0,2\pi)\\
y=\frac{1}{2}\cdot x\cdot \cos x \\
{y}'=({\frac{1}{2}\cdot x\cdot \cos x})' \\
{y}'=\frac{1}{2}\cdot ({x\cdot \cos x})' \\
{y}'=\frac{1}{2}\cdot ({x}'\cdot \cos x+x\cdot (\cos x)') \\
{y}'=\frac{1}{2}\cdot(\cos x-x\cdot \sin x) \\
{y}'=0 \\
\frac{1}{2}\cdot(\cos x-x\cdot \sin x)=0 \\ $
$\hspace{1cm}$
$\\
\cos x-x\cdot \sin x=0\ /(\cos x) \\
1-x\cdot \tan x=0 \\
-x\cdot \tan x=-1 \\
x\cdot \tan x=1 \\
x=\frac{1}{\tan x} \\
x=\cot x $

Hint:
The roots are close to the vertical asymptotes of the cotangent, occurring at $x=k\pi$.
To get a first approximation, we can linearize the cotangent close to a root and 
$$\cot x-x\approx\frac1{x-k\pi}-x=0.$$
The positive solution is 
$$x=\frac{k\pi+\sqrt{k^2\pi^2+4}}{2}.$$
From this, you start Newton's iterations, which will quickly converge.
Though a bit cluttered, the plot below shows you how the hyperbolas match the true curve close to the roots:

