Answer
$$\sin x=-\sqrt{1-\cos^2x}$$
This is a true statement only when the angle $x$ lies in quadrant III or IV.
Work Step by Step
$$\sin x=-\sqrt{1-\cos^2x}$$
As from a Pythagorean Identity: $$\sin^2x=1-\cos^2x$$
That means $$\sqrt{\sin^2x}=\sqrt{1-\cos^2x}$$
$$|\sin x|=\sqrt{1-\cos^2x}$$
In other words, $$\sin x=\pm\sqrt{1-\cos^2x}$$
So we see now that there are 2 cases of $\sin x$, whether it is going to be $\sin x=\sqrt{1-\cos^2x}$ when $\sin x\ge0$ or $\sin x=-\sqrt{1-\cos^2x}$ when $\sin x\lt0$.
In the requirement of the exercise given, it asks for the situation when $$\sin x=-\sqrt{1-\cos^2x}$$which happens when $$\sin x\lt0$$
That happens when the angle $x$ lies in quadrant III or IV.