If $f\left(x\right)=sgn(\cos 2x-2\sin x+3)$, where $sgn\left(\right)$ is the signum function, then $f\left(x\right)$

Given function is $f\left(x\right)=sgn(\cos 2x-2\sin x+3)$,

Since $\cos 2x=1-2{\sin }^{2}x$

$\Rightarrow f\left(x\right)=\left(1-2{\sin }^{2}x-2\sin x3\right)$

$\Rightarrow f\left(x\right)=\left(4-2{\sin }^{2}x-2\sin x\right)$

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\mathrm{sgn}\left[2(2-\mathrm{sinx})(1-\mathrm{sinx})\right]$

Assume $g\left(x\right)=\left\{2\right(2-\sin x\left)\right(1-\sin x\left)\right\}$,

then $g\left(x\right)\ge 0$, since $-1\le \sin x\le 1$

And $(1-\sin x)=0$ when  $x=(2n+1)\frac{\pi }{2},n\in I$

Then, $f\left(x\right)=\left\{\begin{array}{l}0,x=(2n+1)\frac{\pi }{2}\\ 1, x\ne (2n+1)\frac{\pi }{2}\end{array}\right\}$

Then, it can be said that $f\left(x\right)$ is discontinuous at $x=(2n+1)\frac{\pi }{2}$ which is an isolated type of discontinuity.

Hence, option-$3$ is the correct answer.