The number of roots of $x^2-2=\left[\sin x\right],$ where [.] stands for the greatest integer function, is

Given equation is $x^2-2=\left[\sin x\right].$

There are three possibilities:

$\left[\sin x\right]=-1,\left[\sin x\right]=0,\left[\sin x\right]=1$

Case I: If $\left[\sin x\right]=-1$

$\begin{array}{l} x^2=1\\ \mathrm{or}x=\pm 1\end{array}$

$\begin{array}{l}\mathrm{When} x=1,\left[\sin x\right]=0.\\ \mathrm{But} x=-1\Rightarrow \left[\sin x\right]=-1\\ x=-1\mathrm{is} \mathrm{a} \mathrm{solution}.\end{array}$

Case II: If $\left[\sin x\right]=0$ the equation is

$\begin{array}{l} x^2=2\\ \mathrm{or} x\pm \sqrt{2}\\ \left[\sin \sqrt{2}\right]=0\\ \mathrm{But} \left[\sin (-\sqrt{2})\right]=-1.\\ \mathrm{So}, x=\sqrt{2} \mathrm{is} \mathrm{a} \mathrm{solution}.\end{array}$

Case III:  If $\left[\sin x\right]=1$

$x=\frac{\pi }{2},\frac{5\pi }{2},\frac{9\pi }{2}$

Clearly, these values do not satisfy the original equation.
Thus, number of solutions : 2