I came across the following question in a past paper for an SDE course I am following: Let $(B_t)_{t\geq 0}$ b on the filtered probability space $(\Omega,\mathcal F,(\mathcal F_t)_{t\geq 0},\mathbb P)$. Consider the SDE
$$dX_t=-\frac{1}{2}X_tdt+\sqrt{1-X_t^2}dB_t,\quad\quad X_0=0. \tag{1}\label{eq1}$$
The question as given has two parts:
a) Find a solution of the form $f(B_t)$ and investigate the properties $f$ should have.
b) Show that a solution exists for $t<\tau:=\inf\{s>0:B_s\notin[-\frac{\pi}{2},\frac{-\pi}{2}]\}$.
Using It$\hat{\text{o}}$'s formula it is easy to derive that we require $f''(x)=-f(x)$ and $f'(x)=\sqrt{1-f(x)^2}$ for all real $x$. The only real valued function I'm aware of satisfying these properties is $\sin$, so we set $f(x)=\sin(x)$ on $\mathbb R$. As we derived $f$ via It$\hat{\text{o}}$'s formula we know then that $X_t=\sin(B_t)$ satisfies \eqref{eq1} (technically we should check it but that is easy to do). I believe this is a satisfactory answer to part a) (please inform me if it is not).
I am not quite sure what is being asked in part b), however. As far as I am aware we can say that $X_t$ is a solution to the SDE \eqref{eq1} on an interval $I:=[0,T]$ if it satisfies the integral form of the SDE for every $t\in I$, which we know $\sin(B_t)$ does, and if it is continuous and $(\mathcal F_t)$ adapted (which $\sin(B_t)$ is), and if $$\int_0^T\frac{1}{2}dt+\int_0^T\sqrt{1-X_t^2}dt<\infty\qquad \text{a.s}.$$ Now as far as I am aware $X_t=\sin(B_t)$ satisfies all 3 of the above listed properties for any $T>0$, so I would just have concluded that SDE \eqref{eq1} has a solution no matter the value of $t$. Question b) however makes me think something funky happens outside of $[-\frac{\pi}{2},\frac{-\pi}{2}]$, which I presume has something to do with $\sin$ being monotone increasing on that interval. Other than that I am clueless as to how to answer b) and I am worried that it implies I have a deep misunderstanding of the subject. Any comments on the question and my thought processes would be welcome.
No, there is no deep misunderstanding. You are simply missing the fact that the identity
$$\cos(x) = \sqrt{1-\sin(x)^2}$$
holds if, and only if, $\cos(x) \geq 0$. In particular,
$$\cos(B_t) = \sqrt{1-X_t^2} = \sqrt{1-\sin(B_t^2)}$$
fails to hold when the Brownian motion exits the interval $[-\pi/2,\pi/2]$.