I have been working on several trigonometric-linear functions in order to get analytical results. In fact, I think there are no algebraical ways to get solutions for them. Perhaps there is need to consider several points of view.

The equations

$\cos x = x$, $\tan x = x$ and $\sin x = x$

Certainly, at least on $\mathbb{R}$ there is:

$\cos x = x$ has got a solution on $x \approx 0.7390851332$.

$\sin x = x$ has got a solution on $x = 0$


$\tan x = x$ has got solutions on $x = 0, x \approx 4.493, 7.725, 10.903, 14.066, \cdots$ something like $x = 0 ~\cup x \approx 1.3515 + n\pi$.

Any idea about a way to define the solution in a analytical way? I have been thinking of the complex definition of $\cos x$ as:

$\cos x = \dfrac{e^{ix}+e^{-ix}}{2}$

In order to attack the whole problem. I don't really know. Any reference you people may give to me will be welcomed.


This is too long for a comment.

Because of their intrinsically transcendental nature, there is no way to get closed form solutions of equations such as $x=\cos(x)$ or $x=\tan(x)$ and numerical methods are required.

The one which is probably the most interesting is $x=\tan(x)$ for which we can approximate the solutions quite accurately. The solutions are more or less $$x_n=(2n+1)\frac \pi 2-\epsilon_n$$ where $\epsilon_n$ is a small positive number.

A simple way would be to consider function $$f(x)=x\cos(x)-\sin(x)$$ and develop it as Taylor series (to third order) around $x_0=(2n+1)\frac \pi 2$. This would lead to $$x^{(1)}_n\approx \frac{1}{2} \sqrt{(2 n+1)\pi ^2-8}$$ We also could use one iteration of Newton method and get $$x^{(2)}_n\approx (2n+1)\frac \pi 2-\frac 2{(2n+1)\pi}$$ For the first roots, this would lead to $$\left( \begin{array}{ccc} n & x^{(1)}_n &x^{(2)}_n \\ 1 & 4.49518 & 4.50018 \\ 2 & 7.72561 & 7.72666 \\ 3 & 10.9042 & 10.9046 \\ 4 & 14.0663 & 14.0664 \\ 5 & 17.2208 & 17.2209 \end{array} \right)$$ which are very close to the numbers given in the post.

More accurate approximations are given here $$x_n=q-\frac{1}{q}-\frac{2}{3q^3}-\frac{13}{15q^5}-\frac{146}{105q^7}+\cdots \qquad \text{with}\qquad q=(2n+1)\frac \pi 2$$


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