Let $f\left(x\right)={\sec }^{-1}\left(\right[1+{\sin }^{2}x\left]\right);$  where [.] denotes greatest integer function. Then the set of points where $f\left(x\right)$ is not continuous is

$f(n\pi +)={\sec }^{-1}1=0$  and $f\left(n\pi -\right)={\sec }^{-1}1=0$  and $f\left(n\pi \right)=0$

$\therefore f$  is continuous at $x=n\pi$

$f\left((2n-1)\frac{\pi }{2}+\right)={\sec }^{-1}1=0$  but $f\left((2n-1)\frac{\pi }{2}\right)={\sec }^{-1}2=\frac{\pi }{3}$

$\therefore f$  is discontinuous at $x=(2n-1)\frac{\pi }{2}$  for all $n\in I$