I am looking at a problem (5.1.iii, pg 93 of the pdf or pg 73 of the book: http://th.if.uj.edu.pl/~gudowska/dydaktyka/Oksendal.pdf)
The function $X_t = \text{sin }B_t$ for any $B_0 = a \in (-\pi/2, \pi/2)$ solves the Ito SDE $dX_t = -\frac{1}{2}X_t dt + \sqrt{1 - X_t^2}dBt$ which seems pretty clear from Ito's formula and the identity $sin^2x + cos^2 x = 1$. But the problem stipulates that this solution is only valid for $t < \text{inf}(s > 0 | B_s \not\in[-\pi/2, \pi/2])$.
What is the reason for bounding $t$ to prevent $B_t$ escaping $[-\pi/2, \pi/2]$? If the trig functions are periodically extended, shouldn't this solution be valid for all $t$?
Directly outside of that interval the cosine is negative, the square root would need to change its sign. Thus the given function would not solve the given SDE. It does not necessarily mean that the SDE does not have a solution, but it will no longer be the sine.