Find Range of $a$ if $ \lfloor \frac{1}{3}+2a \sin ^3 x \rfloor $ is an Even function

Question

Find Range of $a$ if $$ f(x)= \lfloor \frac{1}{3}+2a \sin ^3 x \rfloor $$ is an Even function

My try:

we have $$f(-x)=f(x)$$ $\implies$

$$ \lfloor \frac{1}{3}+2a \sin ^3 x \rfloor=\lfloor \frac{1}{3}-2a \sin ^3 x \rfloor \tag{1}$$

Obviously $a=0$ is the possibility

Now how can we determine other values of $a$ which makes $(1)$ True?

Answer 1

Obviously it is symmetric in $a$ and trivial for $a=0$, so we can just examine $a>0$:

Suppose $2a\leq \tfrac13$, then $$\bigg\lfloor\frac13+\underbrace{2a\sin^3x}_{\in [-\tfrac13,\tfrac13]}\bigg\rfloor = 0 = \bigg\lfloor\frac13-\underbrace{2a\sin^3x}_{\in [-\tfrac13,\tfrac13]}\bigg\rfloor$$

Suppose $2a>\tfrac13$, then at $x = \tfrac\pi2$

$$\bigg\lfloor\frac13+2a\sin^3 \tfrac\pi2\bigg\rfloor = \bigg\lfloor\frac13+2a\bigg\rfloor \geq 0,$$ but $$\bigg\lfloor\frac13-2a\sin^3 \tfrac\pi2\bigg\rfloor = \bigg\lfloor\frac13 -2a\bigg\rfloor < 0.$$ Hence $a \in [-\tfrac16,\tfrac16]$.

Answer 2

We need $f\left(-\frac{\pi}{2}\right)=f\left(+\frac{\pi}{2}\right)$. Therefore, $$\left\lfloor \frac{1}{3}+2a\right\rfloor =\left\lfloor\frac{1}{3}-2a\right\rfloor\,.$$ If $|a|> \dfrac{1}{6}$, then one of the numbers $\frac{1}{3}-2a$ and $\frac{1}{3}+2a$ is negative, while the other is positive, so the equality above does not happen. What happens if $|a|\leq\dfrac{1}{6}$?

Answer 3

Let $z = 2a \sin^3 x$, then we need to find when $ \left\lfloor \frac 13 - z \right\rfloor = \left\lfloor \frac 13 + z \right\rfloor $?

We suppose first that $z$ is an integer. Then we get

\begin{align} \left\lfloor \frac 13 - z \right\rfloor &= \left\lfloor \frac 13 + z \right\rfloor \\ \left\lfloor \frac 13 \right\rfloor - z &= \left\lfloor \frac 13 \right\rfloor + z \\ z &= 0 \end{align}

So, if $z$ is not an integer, we can suppose that $0 < z < 1$.

Hence $ \left\lfloor \frac 13 - z \right\rfloor = \begin{cases} 0 & \text{If $0 < z \le \dfrac 13$} \\ -1 & \text{If $\dfrac 13 < z < 1$} \\ \end{cases} $

and $ \left\lfloor \frac 13 + z \right\rfloor = \begin{cases} 0 & \text{If $0 < z < \dfrac 23$} \\ 1 & \text{If $\dfrac 23 < z < 1$} \\ \end{cases} $

We can conclude that we need

\begin{align} |z| &\le \dfrac 13 \\ |2a \sin^3 x| &\le \dfrac 13 \\ 2|a| &\le \dfrac 13 &\text{(Since $\max |\sin^3 x| = 1$}) \\ |a| &\le \dfrac 16 \end{align}