Calculate time when star is above altitude 30°
Yes, this is the correct approach.  The $h$ in the equation is the altitude above the horizon of the object at which you consider it to be rising or setting.  This is typically non-zero, because of atmospheric refraction, and, in the case of the Sun or the Moon, because of their finite diameters.  In your case, the object 'rises' when it climbes above $h=30^\circ$ and 'sets' when it drops below that altitude.

If $\left|\cos(LHA)\right|&gt;1$, there is no solution, because your object never crosses the $h=30^\circ$ line.  $LHA$ is the local hour angle, and you can find the local sidereal time $\theta$ using

$LHA = \theta - \alpha,$

where $\alpha$ is the right ascension of your object.