Given
$$ t x' = (1 - 4t^2) \tan x $$
Easily, by separation of variables, we obtain
$$ x(t) = \arcsin( C t e^{-2t^2}) $$
However, $x = 0, \pi/2, ... $ are also solution, equilibirum solutions. This is the answer given in the book which I think should also include the equilibrium solutions as well. Maybe, do we need to specify the interval where $x$ lies to avoid having equilibrium solutions?
Observe that $x(t)=\pi/2$ is not a solution, since the right hand side is not defined. The equilibrium solutions are $x(t)=k\,\pi$, $k\in\mathbb{Z}$. They are obtained from the general solution by letting $C=0$.