1000 questions! Looks nice

This is a tricky problem.

It is given that $f(x) = \dfrac {2x(\sin{x}+\tan{x})}{2 \lfloor 2+\frac{x}{\pi}\rfloor -3} = \dfrac {g(x)}{h(x)}$

We note that:

• $\sin{x}$ is odd, $\tan{x}$ is odd and the sum of two odd functions is odd $\Rightarrow (\sin{x}+\tan{x})$ is odd.
• $2x$ is odd, $(\sin{x}+\tan{x})$ is odd and the product of two odd functions is even $\Rightarrow g(x) = 2x(\sin{x}+\tan{x})$ is even.
• We note that $h(x) = 2 \lfloor 2+\frac{x}{\pi}\rfloor -3$ is strictly not an odd function because $h (-n\pi) = -h(n\pi)-2$ , where $n$ is an integer but for $x \ne n\pi$ , $h(-x) = -h(x)$ , it is an odd function.
• But we find that $g(n \pi)=0$ , therefore $f(x)$ is, in effect, a quotient of an even function $g(x)$ and an odd function $h(x\ne n \pi)$ , which is an odd function.

Therefore, $\boxed{f(x) \text{ is an odd function}}$ .

This is how it looks like. Isn't it odd?!!!