# Solution for the trigonometric-linear function

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 $x = 0$

and

$\tan x = x$ 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.

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