If the function $$f(x)=[4.8+a\sin x]$$ where $[.]$ denotes the greatest integer function, is an even function then find the range of $a$.
So, if the function of even then $f(-x)= f(x)$.
Therefore,$$[4.8+a\sin x]=[4.8-a\sin x]$$ $$\implies [0.8+a\sin x]=[0.8-a\sin x]$$ But I don't have any idea what to do next, please help. Cheers
We want that $$\lfloor 0.8+a y\rfloor=\lfloor 0.8-a y\rfloor\quad(-1\leq y\leq 1)\ .$$ If $|a|<0.2$ then $$\lfloor 0.8+a y\rfloor=\lfloor 0.8-a y\rfloor=0\quad(-1\leq y\leq 1)\ .$$ On the other hand, if $|a|\geq0.2$, choose $y:={\displaystyle{0.2\over a}}\in[{-1},1]$ and obtain $$\lfloor 0.8+a y\rfloor=1,\qquad \lfloor 0.8-a y\rfloor=0\ .$$ It follows that the admissible range for $a$ in this problem is $\>]{-0.2}, 0.2[\>$.