Calculus en tan function

Question

I got this easy problem but I do not know how to face it. I tried to get the tangent from both side but I got nowhere. I tried to use some tangent properties but then again I missed the way.

Prove that $\tan(x)=x$ has infinitely many solutions.

Answer 1

We know that in every interval $I_{k} = ]k \frac{\pi}{2}, (k +1)\frac{\pi}{2}[$ the function $f(x) = \text{tan}(x)$ is continuous and ranges from $-\infty$ to $\infty$. In the same interval the function $g(x) = x$ ranges from $-k\frac{\pi}{2}$ to $(k + 1)\frac{\pi}{2}$ so there are two numbers $x_{1}, x_{2} \in I_{k}$ such that $x_{1} < x_{2}$, $f(x_{1}) < g(x_{1})$ and $f(x_{2}) > g(x_{2})$.

If we define the new function $h(x) = f(x) - g(x)$ this means that $h(x_{1}) < 0$ and $h(x_{2}) > 0$, since $h$ is continuous in $I_{k}$ (difference of continous functions) there exists $y \in I_{k}$ such that $h(y) = 0$ (We're using the intermediate value theorem). From that it follows $f(y) = g(y)$.

Since we have found such an $y$ for every interval $I_{k}$ we have proved that the equation $\text{tan}(x) = x$ has infinitely many solutions.

Answer 2

At $x=n\pi, \tan x = 0 < x.$ Now $\lim_{x\to (n\pi +\pi/2)+} = \infty$ so for some value $n\pi < x<n\pi+\pi/2, x<\tan x$ Since $f(x)=x-\tan x$ is continuous on $(n\pi, n\pi+\pi/2)$ the is a point $x$ in this interval where $x=\tan x$

Answer 3

Note that the graph of $$y=\tan (x)$$ has infinitely many vertical asymptotes and between any two of these asymptotes the graph covers the entire $$y=(-\infty, \infty)$$

Thus between any two consecutive asymptotes the graph of $y=x$ meets the graph of $y=\tan(x)$

That is infinitely many intersections.

Answer 4

Just observe that

• $\forall k\in \mathbb{Z}, \, \tan x$ for $x\in I_k=(-\pi/2+k\pi,\pi/2+k\pi)$ is continuos, surjective and strictly increasing
• then $\forall k\in \mathbb{Z},\,f(x)=\tan x-x$ intersect $x$ axis exactly once

thus $\tan x = x$ has infinitely many solutions.