Find all real $x$ such that $1990[x] +1989[-x]=1$ (where $[x]$ is the floor function for $x$).
My effort
Rearranging our equation we have :
\begin{array}{c} 1990[x]+1989[-x]&=1 \\ 1989([x]+[-x])+[x] &=1 \\ \end{array}
Supposing that $x$ is an integer ,I have that $[x]+[-x]=0$ and the problem breaks down to $$[x]=1$$ which has the only solution $x=1$
Else ,$x$ is a real number with nonzero fractional part and $[x]+[-x]=-1$ which yields in our case
\begin{array}{c} -1989 + [x] &= 1 \\ [x] &=1990 \\ \end{array}
For this to happen we must therefore have that $x \in (1990,1991)$
Question
Is my effort complete and correct ?What would have been other ways to approach the problem ?