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