Show that the equation $\tan x = −x$ has a single root on the interval $(\frac{-\pi}{2}, \frac{\pi}{2})$. Show that the equation $\tan x = −x$ has a single root on the interval $(\frac{-\pi}{2}, \frac{\pi}{2})$.
We have $0 \in (\frac{-\pi}{2}, \frac{\pi}{2})$ then $\tan(0)=-0$
But its unique because $f(x)=\tan(x)+x$ is differentiable in $(\frac{-\pi}{2}, \frac{\pi}{2})$ then $f'(x)= 1+\frac{1}{\cos^2(x)}=0$ we have $-1=\cos^{2}(x)$ its a contradiction. Am I right?

Alternative approach:
Assuming that $-\pi/2 < x < \pi/2$ and that $x \neq 0$:

*

*$x$ is positive if and only if $\tan(x)$ is positive.

*$x$ is negative if and only if $\tan(x)$ is negative.

The above two bullet points follow, since the $\cos(x)$ is positive throughout the interval, the $\sin(x)$ is positive for $x$ positive and the $\sin(x)$ is negative for $x$ negative.
Therefore, $x = 0$ is the only possible solution to the original problem.
