Answer
$\frac{3\pi}{2}~rad \lt \phi \lt 2\pi~rad$
Work Step by Step
$x = x_m~cos(\omega~t+\phi)$
When $x = 0$ and the object is moving in the positive direction, then $\phi = \frac{3\pi}{2}~rad$
When $x = x_m$, then $\phi = 2\pi~rad$
Since the object is between these two positions, then $~~\frac{3\pi}{2}~rad \lt \phi \lt 2\pi~rad$
