Problem Statement:-
Solve: $$\dfrac{1}{\left\lfloor{x}\right\rfloor}+\dfrac{1}{\left\lfloor{2x}\right\rfloor}=\{x\}+\dfrac{1}{3}$$ where $\left\lfloor{x}\right\rfloor$ denotes the integral part of $x$ and $\{x\}=x-\left\lfloor{x}\right\rfloor$
The floor function just eats me up whole whenever I encounter it. I have tried some things to solve it but am just not able to come up with anything useful at all. Here are the things I have tried.
First of all lets define the domain of $x$ which is $x\not\in[0,1)$, because if so happens then $\left\lfloor{x}\right\rfloor=0$ which should not happen for $\dfrac{1}{\left\lfloor{x}\right\rfloor}$ and $\dfrac{1}{\left\lfloor{2x}\right\rfloor}$ to be defined.
After this I was not able to come up with a good line of thought for attempting the problem so, please guide me as to what should be my line of thought for attempting problems related to the integral part of $x$.