Solutions of $\tan(x)=-\lambda x$? Motivated by a physical problem I would be interested in the solutions of $$\tan(x)=-\lambda x$$ with $\lambda,x \in \Bbb R^+$. Especially the first non-trivial solution (the trivial is $x=0$) would be of interest.
The problem with $\lambda=-1$ has some intriguing solutions, discussed here. So maybe there is a chance for a nice solution here as well? Nice shall not mean non-transcendental.
The methods from this link could maybe be modified?
 A: Most has been already said in comments and answers.
If we think about approximations, we could notice that the function can be quite well represented by its $[2n,2n]$ Padé approximant built at $x=0$. For example, a simple one could be
$$-\frac{\tan (x)}{x}\sim \frac{-1+\frac{1}{9}x^2-\frac{1}{945}x^4 } {1-\frac{4 }{9}x^2+\frac{1}{63}x^4 }$$ which shows an absolute error of $5.38\times 10^{-5}$ for $x=\frac{7\pi}{16}$ (using the next approximant, this error would become $7.24\times 10^{-9}$).
Then the approximation of the first root of $-\frac{\tan (x)}{x}=\lambda$ reduces to a quadratic (or cubic) equation in $x^2$.
Edit
If you have a look at this question of mine,we could work the problem slightly better considering the $[4,2]$ Padé approximant of
$$\left(x-\frac{\pi }{2}\right) \left(x+\frac{\pi }{2}\right)\frac{ \tan (x)}{x}$$
This would lead to
$$\frac{ \tan (x)}{x}=\frac{a_0+a_1x^2+a_2 x^4}{\left(x-\frac{\pi }{2}\right) \left(x+\frac{\pi }{2}\right) \left(1+a_3 x^2\right)}$$ where
$$a_0=-\frac{\pi ^2}{4}\qquad a_1=\frac{-1680+140 \pi ^2+3 \pi ^4}{168 \left(\pi ^2-10\right)}$$
$$a_2=\frac{1680-180 \pi ^2+\pi ^4}{2520 \left(\pi ^2-10\right)}\qquad a_3=\frac{168-17 \pi ^2}{42 \left(\pi ^2-10\right)}$$ which is significantly better.
A: For any $x \neq 0$, this is equivalent to $\lambda = \frac{-\tan(x)}{x}$. You can see a plot of this here. (For $x=0$, of course, it's satisfied by any $\lambda$ whatsoever.) But this probably isn't what you're asking.
For a given value of $\lambda$, on the other hand, this is significantly harder to solve. Dave L. Renfro in the comments points to page thirteen of these slides, which suggests that there is no "nice" non-trivial solution to $\tan(x) = R(x)$ for any nonzero rational function $R$. Unfortunately, $-\lambda x$ for any $\lambda \neq 0$ counts as a nonzero rational function.
This depends on Schanuel's conjecture, which hasn't been proven—but it's also unlikely to be disproven in a StackExchange answer. "Nice" here means "made of rational numbers and elementary functions".
A: $$tan(x)=-\lambda x$$
$$\frac{tan(x)}{x}=-\lambda,\ \ x\neq 0$$
1.)
To solve this equation, you could use the partial inverses (= the branches of the inverse relation) of the elementary function $x\mapsto \frac{tan(x)}{x}$. It follows with Ritt's theorem on elementary inverse functions (1925) that the function doesn't have elementary partial inverses if its domain contains an open set. That means, it is not possible to solve the equation by only applying only elementary inverse operations of the elementary operations contained in the equation terms.
2.)
See this reference: Kheyfits, A. I.: Explicit solutions of transcendental equations and the Lambert W function. The author writes: "The method can be also applied to many other equations, like
... $w\ \tan w=z$ ..."
Apply this method to $\frac{tan(x)}{x}$, if this is possible.
Or use Lagrange inversion.
3.)
Let $T$ denote the inverse of the function $x\mapsto \frac{tan(x)}{x}$. Applying this to your equation above yields the solution $x=T(-\lambda)$.
4.)
For a given $\lambda$, you can use $\tan(x)+\lambda x=0$. The Taylor series of the function $x\mapsto \tan(x)+\lambda x$ is more simple than that of the function $x\mapsto \frac{\tan(x)}{x}$, and therefore its Lagrange inversion.
5.)
There is no elementary solution that you can read only from the equation. If you use a Special function or a series representation, e.g. a Taylor series, you have to calculate single values. The simplest way is therefore to calculate single values without using a formula for the partial inverses. Calculate a table of values of $x$ and $\frac{\tan(x)}{x}$ and invert this function. And apply the partial inverse functions to your $-\lambda$ as above.
6.)
Wolfram Alpha gives some numerical solutions. See e.g. here.
A: Here is an outline how the problem can be tackled using perturbation theory on a quantum mechanical problem, that is directly related to it.
This is a graphical representation how the perturbation series approaches the exact curve for the first non-trivial non-negative root of (with swapped axis) $$\tan y = -x y$$ (this is the blue "exact" curve) the perturbation series is
$$ u(c) = 1+c-c^2\frac{1}{4}+c^3\left(\frac{1}{16}-\frac{\pi^2-9}{48}\right)+c^4\left(-\frac{1}{64} +\frac{\pi^2-9}{192}-\frac{\pi^2-9}{96}+\frac{\pi^2-10}{64}\right) + \mathcal{O}(c^5) $$
The plot shows $$ y(x) = \frac{\pi}{2}\sqrt{u(c\left(\frac{8}{x\pi^2}\right))} $$ to order $n$ in $c^n$ (yellow: $1$ , green $2$, red $3$, purple $4$) and the exact curve $\tan y = xy$ (blue)
$$ y(x) = \frac{\pi}{2}\sqrt{u(c\left(\frac{8}{x\pi^2}\right))} $$ to order $n$ in $c^n$ (yellow: $1$ , green $2$, red $3$, purple $4$) and the exact curve $\tan y = xy$ (blue)">
