What are the solutions of $x=\cot x$?

Question

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 $

Answer 1

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:

Answer 2

It is a transcendental equation and you will have to find the solution numerically.

Answer 3

Considering instead the function $$f(x)=\cos(x)-x\sin(x)$$

its solutions can be very well approximated developing $f(x)$ as a Taylor series built around $x=n \pi$ and using later series reversion. This gives for $n>0$ $$x_{n+1}=q+\frac{1}{q}-\frac{4}{3 q^3}+\frac{53}{15 q^5}-\frac{1226}{105 q^7}+\frac{13597}{315 q^9}-\frac{65782}{385 q^{11}}+O\left(\frac{1}{q^{13}}\right)$$ where $\color{red}{q=n \pi}$.

$$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 1 & 3.42544796823 & 3.42561845948 \\ 2 & 6.43729815231 & 6.43729817917 \\ 3 & 9.52933440522 & 9.52933440536 \\ 4 & 12.6452872239 & 12.6452872239 \\ 5 & 15.7712848748 & 15.7712848748 \end{array} \right)$$

Concerning the first root, we can approximate it building the $[2,2p]$ Padé approximant and, depending on $p$, the square of the approximate solution will correspond to the sequence $$\left\{\frac{2}{3},\frac{36}{49},\frac{1470}{1987},\frac{111272}{150337},\frac{1353 0330}{18279971},\frac{2412956172}{3259985969},\frac{593317446358}{801591982115} \right\}$$