Solutions to $\frac1{\lfloor x\rfloor}+\frac1{\lfloor 2x\rfloor}=\{x\}+\frac13$ 
Find all solutions to $$\dfrac{1}{\lfloor x\rfloor}+\dfrac{1}{\lfloor 2x\rfloor}=\{x\}+\dfrac{1}{3}$$

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Unfortunately I have no idea as to how to go about this. On rearranging, I got $$3\lfloor 2x\rfloor = 3\lfloor x\rfloor\{x\}-2\lfloor x\rfloor$$
I'm not sure about what to do with the $3\lfloor 2x\rfloor $ term; I'd prefer to resolve it in terms of $\lfloor x\rfloor $ but am not able to. All that struck me was using the identity for $\lfloor nx\rfloor, n\in \Bbb Z$. However on first glance, it did not strike me as particularly useful.$$$$
I would be grateful for any help. Many thanks! 

$$\dfrac{1}{\lfloor x\rfloor}+\dfrac{1}{\lfloor 2x\rfloor}=\{x\}+\dfrac{1}{3}\tag1$$
We have $\lfloor x\rfloor$ and $\lfloor 2x\rfloor$, so one way is to separate it into two cases :
Case 1 : $x=n+\alpha$ where $n\not=0\in\mathbb Z,0\le\alpha\lt 1/2$
Case 2 : $x=n+\alpha$ where $n\not=0\in\mathbb Z,1/2\le\alpha\lt 1$
For case 1, $$\begin{align}(1)&\implies \frac 1n+\frac{1}{2n}=\alpha+\frac 13\\&\implies  \alpha=\frac{9-2n}{6n}\\&\implies0\le \frac{9-2n}{6n}\lt \frac 12\\&\implies (n,\alpha)=(2,5/12),(3,1/6),(4,1/24)\end{align}$$
I think that you can do for case 2 similarly.
