solving $\frac{x}{3}+{[\frac{x}{3}]} = \sin(x) + [\sin(x)]$ for real x , in an efficient manner

Question

$$\frac{x}{3}+{\left[\frac{x}{3}\right]}=\sin(x)+[\sin(x)]$$ I know the answer and the solution. $$-1 \le \sin(x) \le 1\Rightarrow-2\le\sin(x)+[\sin(x)]\le 2 \Rightarrow -2 \le \frac{x}{3}+\left[\frac{x}{3}\right] \le 2 $$ now there are different situations :

$x=3$ is unacceptable $$0 \le x \lt 3 \Rightarrow0 \le \frac{x}{3} \lt 1\Rightarrow \left[\frac{x}{3}\right]=0 \land[\sin(x)]=0\Rightarrow\frac{x}{3}=\sin(x)$$ which has 2 acceptable answers . $$-3 \le x \lt 0 \Rightarrow-1 \le \frac{x}{3}\lt 0\Rightarrow \left[\frac{x}{3}\right] =-1\land [\sin(x)] = -1\Rightarrow\frac{x}{3} =\sin (x) $$ which has one acceptable answer. so the equation has 3 real answers . I want to know if there is a better way to solve it. ( beautiful equation , isn't it?!)

Answer 1

There is a more holistic way to solve this. The function $y\mapsto y+[y]$ is strictly increasing; in particular, it is one-to-one. Therefore $y+[y]=z+[z]$ if and only if $y=z$. So you can deduce right away that you're looking exactly for solutions to $\frac x3=\sin x$ (which is what you eventually found).