Use the intermediate value theorem to show that the equation,

$tan(x)=2x$

has an infinite amount of real solutions.

So far I have used the IVT to show that for $f(x)=tan(x)$ in the interval $(-\frac{\pi}{2},\frac{\pi}{2})$ there is a $L$ between $-\mathrm{\infty}$ and $\mathrm{\infty}$ such that there is a value $c$ where $f(c)=L$

I also have shown that for $g(x)=2x$ in the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$ there is a $K$ between $-\pi $ and $\pi $ such that there is a value $d$ where $f(d)=K$

Would it be correct to say that because each of these functions have a value for each point in the interval and that the range of $f(x)$ consists of all reals that $f(x)$ must intersect with $g(x)$ or there is some value $x$ in the interval such that $2x=tan(x)$? If this is correct then how would I extend this result to all intervals $[\pi (n-\frac{1}{2}),\pi (n+\frac{1}{2})]$?

$tan(x)=2x$

has an infinite amount of real solutions.

So far I have used the IVT to show that for $f(x)=tan(x)$ in the interval $(-\frac{\pi}{2},\frac{\pi}{2})$ there is a $L$ between $-\mathrm{\infty}$ and $\mathrm{\infty}$ such that there is a value $c$ where $f(c)=L$

I also have shown that for $g(x)=2x$ in the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$ there is a $K$ between $-\pi $ and $\pi $ such that there is a value $d$ where $f(d)=K$

Would it be correct to say that because each of these functions have a value for each point in the interval and that the range of $f(x)$ consists of all reals that $f(x)$ must intersect with $g(x)$ or there is some value $x$ in the interval such that $2x=tan(x)$? If this is correct then how would I extend this result to all intervals $[\pi (n-\frac{1}{2}),\pi (n+\frac{1}{2})]$?