How solve this: if $\sin{(ax+b)}=\sin{29x}$ for all integers $x$, find the smallest possible value of $a$ Question:

Assume that $a$ and $b$ are nonnegative real numbers such that $$\sin{(ax+b)}=\sin{29x}$$ for all integers $x$.
Find
the smallest possible value of $a$.

This problem  is from a China book, and the book  only have result, the answer is  $a_{\min}=10\pi-29$,
I fell this reslut is surprising, can you help me explain why?
Thank you
 A: Because $\sin (x)$ has a period of $2\pi$, $\sin(\alpha) = \sin(\alpha + 2k\pi)$ for any integer $k$. Also, $\sin(x)$ is an odd function, so $\sin(-\alpha) = -\sin(\alpha)$.
Applying this to your problem, we see that $\sin(\pm 29x + 2k\pi) = \sin(29x)$, where in this case $a=29$ and $b=2k\pi$ and $k$ is an integer. But we want to minimize the value of $a$. Since $x$ is an integer, subtracting multiples of $2\pi x$ won't change the value of the expression. So $\sin(29x) = \sin(29x - 2\pi x) = \cdots =  \sin(29x - 10\pi x) = \sin((29-10\pi)x)$. However, $(29-10\pi)$ is negative so we need to take the absolute value of this quantity to obtain $-\sin(29x) = \sin((10\pi - 29)x)$, because $\sin(x)$ is odd. But now we have $-\sin(29x)$ instead of $\sin(29x)$. To flip the sign of $-\sin(29x)$, add $\pi$, which is one half of the period of the sine function
So the final answer is $\sin(29x) = \sin((10\pi - 29)x + \pi)$.
As for why the $10\pi$ in particular...that's just arithmetic. You want to make sure that $a$ is minimized, which is the same as making sure that $|29-2k\pi|$ is minimized. For $k=4$, if you plug it into a calculator, this quantity is not as small as it is for $k=5$, and $k=6$ is once again bigger. So you know that $k=5$ and thus $10 \pi - 29$ is the smallest value of $a$.
A: Thornshadow17432's answer is almost right. But $\sin((10π−29)x)$ equals $-\sin(29x)$, not $+\sin(29x)$. To rectify this, we must bring $b$ in:
$$\sin((10π−29)x + \pi) = \sin(29x)$$
