Let $f(x)=\cos x-1$
Here $f(x)$ is continuous on $[0,2\pi]$ ans also differentiable at $(0,2\pi)$.
Hence all the conditions of Rolle's theorem are satisfied.
There exists at least one point $c\in (0,2\pi)$ for which $f'(c)=0$
$\Rightarrow -\sin c=0$
Therefore $f(c)=\cos \pi-1$
Therefore by geometrical interpretation of Rolle's theorem $(\pi,-2)$ is a point on $y=\cos x-1$ where the tangent is parallel to $x$-axis.