How many fixed points of $\tan x$ are there if $x>0$, $x\neq k \pi + \frac{\pi}{2}$

Question

Is the following statement is true or false

Consider the function $\tan x $ on the set $S=\{ x \in \mathbb{R} : x \ge 0 , x \neq k \pi + \frac{\pi}{2} \text{ for any } k \in \mathbb{N} \cup \{0\} \}$ We say that it has a fixed point in $S$ if there exist $x \in S$ such that $\tan x = x$.
Then there are infinitely many fixed points

My attempt : I think this statement is false because the only point that satisfied $\tan x =x$ is $0$ .

$\implies \tan 0 =0 $

Therfore there is a unique fixed point.

Answer 1

Consider function $f(x)=\tan x - x$. Let $k\in\Bbb N_{>0}$.

• $f(k\pi)=-x< 0$
• $\lim_{x\to {k\pi+\frac\pi2}^-}=\infty>0$ since $\lim_{x\to {k\pi+\frac\pi2}^-}\tan x=\infty$

By the intermedia value theorem, since both $\tan(x)$ and $x$ are continuous on the interval $[k\pi, k\pi+\frac\pi2)$, there is value $x_k\in (k\pi, k\pi+\frac\pi2)$ such that $f(x_k)=0$.

$x_1, x_2, \cdots$ are all distinct since $x_1<2\pi<x_2<3\pi<x_3\cdots$.
Hence there are infinitely many fixed points in $S$, $x_1, x_2, \cdots.$