$f(x) = \text{sgn}(\cos 2x - 2 \sin x + 3)$, where sgn(.) is the signum function. Then $f(x)$
A) is continuous over its domain.
B) has a missing point discontinuity.
C) has an isolated point discontinuity.
D) irremovable discontinuity.
The quadratic equation can be written as:
$-2\sin^2 x - 2\sin x + 4$
Making perfect square,
$-2\Big(\sin x+\frac{1}{2}\Big)^2+\frac{9}{2}$
Now, this would always be positive. Its range lying between $\Big(0,\frac{9}{2}\Big)$. Therefore, its signum would always be $1$. So, the answer according to me should be A). But the answer is B), D).
Any help would be appreciated.
Consider the expression $$-2\Big(\sin x+\frac{1}{2}\Big)^2+\frac{9}{2}$$ This is not always positive, since the maximum value of $\sin(x)$ is $1$, and evaluated at $\sin(x)=1$, this is equal to $$-2\Big(\frac{3}{2}\Big)^2+\frac{9}{2}=0$$ and $\text{sgn}(0)=0$. Thus, your original function is discontinuous at $$x=\frac{(4n+1)\pi}{2}$$ for $n\in\mathbb Z^+$.