I am going through exercises in Edwards + Penney's Elementary Differential Equations textbook. I am confused for the solution in the back for question 1.3.30.
The question states:
Determine in terms of $a$ and $b$ how many different solutions the initial value problem $y\prime = - \sqrt{1-y^2}, y(a)=b$.
The solution states: The initial value problem $y\prime = - \sqrt{1-y^2}, y(a)=b$ has a unique solution if $|b|<1$; no solution if $|b|>1$ and infinitely many solutions (defined for all $x$) if $b=\pm1$
Now since, after using separation of variables we get: $arcsin(y)= x + C$ it clearly makes sense that there are no solutions when $|b|>1$.
I am assuming that when the textbook says there infinitely many solutions when $b=\pm1$ they're referring to the fact that $sin(x)$ is periodic and therefore there are (countably) many solutions of the form $k{\pi}\over{2}$ where $k \in \mathbb{Z}$. But isn't this also true for $|b|<1$? For example, if $b= {{1}\over{\sqrt{2}}}<1$, then $arcsin({{1}\over{\sqrt{2}}})= {k{\pi}\over{4}}$ where $k \in \mathbb{Z}$.
What am I missing?