Solving equation $2t=-\tan(1/t)$

Question

I am trying to solve the equation $2t = -\tan(1/t)$, where $t \in (0,1)$. Is there going to be an infinite or finite number of answers?

Answer 1

As you can see, good luck :-D $\ $

Answer 2

Letting $y = 1/t$, $2t = -\tan(1/t) $ becomes $-2 = y \tan(y)$ with $y > 1$.

Since $\tan(k\pi) = 0$, let's look at $z_k =(k\pi-c)\tan(k\pi-c_k)$ where $c_k$ is small.

$\begin{array}\\ z_k &=(k\pi-c_k)\tan(k\pi-c_k)\\ &=(k\pi-c_k)\tan(-c_k)\\ &=-(k\pi-c_k)\tan(c_k)\\ &=-(k\pi-c_k)(c_k+O(c_k^3))\\ &=-k\pi c_k+c^2+O(kc_k^3))\\ \end{array} $

So if $c_k \approx \frac{2}{k\pi} $, then $z_k \approx -2 $.

By continuity and a little fudging, there is a $d_k \approx c_k$ such that $(k\pi-d_k)\tan(k\pi-d_k) = -2 $ so that $\dfrac{2}{k\pi-d_k} =-\tan\left(\dfrac1{k\pi-d_k}\right) $.

So there are an infinite number of solutions, which cluster around zero.

Answer 3

Let $x=1/t$. The equation $\tan(x)=-2/x$ obviously has infinitely many solutions.