If $\sqrt{3}\sin x+\cos x-2={\left(y-1\right)}^2$ for $0\le x\le 8\mathrm{\pi ,}$ then the number of values of the pair $\left(x,y\right)$ is equal to

$\max .\left(\sqrt{3}\sin x+\cos x\right)=\sqrt{{\sqrt{3}}^2+1^2}=2$
and ${\left(y-1\right)}^2\ge 0$
So, LHS $\le 0$ and RHS $\ge 0$
So, the possibility of equality is only when
LHS $=$ RHS $=0$
RHS $=0\mathrm{}\Rightarrow y=1$
LHS $=0\mathrm{}\Rightarrow \sqrt{3}\sin x+\cos x=2$
$\Rightarrow \frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x=1$
$\Rightarrow \cos \left(x-\frac{\pi }{3}\right)=1$
$\Rightarrow x-\frac{\pi }{3}=2n\pi \mathrm{}\Rightarrow x=2n\pi +\frac{\pi }{3}$
$0\le 2n\pi +\frac{\pi }{3}\le 8\pi \mathrm{}\Rightarrow n=0,\mathrm{}1,\mathrm{}2,\mathrm{}3\mathrm{}$
$\Rightarrow \mathrm{}$ There are four values of $x$ and hence four pairs of $\left(x,y\right)$