How would I go about finding the roots of $(e^x-1)-k\arctan(x) = 0$? I have the equation $$(e^x-1)-k\arctan(x) = 0$$ where $0&lt;k \leq \frac 2\pi$ and I was wondering how I would go about starting to determine the amount of real roots of this equation. So far I have just manipulated the equation to get different equations of $x$, however I'm unsure what to do with them.
Current equations for $x$  are  $x = \ln(k\arctan(x)-1)$ and $x = \frac{\tan(e^x-1)}{k}$
Of course $x=0$ is always a solution.  Otherwise, write the equation as $$k = \frac{e^x-1}{\arctan(x)}$$
Call the right side $f(x)$.  The singularity at $x=0$ is removable, with $\lim_{x \to 0} f(x) = 1$.  We also have $\lim_{x \to -\infty} f(x) = 2/\pi$ and $\lim_{x \to +\infty} f(x) = +\infty$.
It appears that $f(x)$ is increasing.  If so, for $2/\pi &lt; k &lt; 1$ and $1 &lt; k &lt; \infty$ there are two real roots ($x=0$ and the root of $f(x)=k$), otherwise there is only $x=0$.
Let $f(x)=e^x-1-k\arctan(x)$.  We have $f(0)=0.$  If $k=0$, obviously $x=0$ is the only solution.
If $k\neq0$, then $f'(x)=e^x-\frac k{1+x^2},$ so that $$f'(x)=0\iff (1+x^2)e^x=k.$$  Since $(1+x^2)e^x$ is increasing, goes to $0$ at $-\infty$ and goes to $\infty$ at $\infty$ we see that $f'$ has exactly one real zero, say $f'(x_0) = 0$.  We see that $f$ decreases on $(-\infty, x_0)$ and increases on $(x_0, \infty)$, so that $f$ attains its minimum at $x_0$.
Since we know that $f(0)=0$, it must be that $x_0&lt;0$ and $x=0$ is the only solution.