Number of solutions of the equation $x^2-2=[\sin x]$, where [.] denotes the greatest integer function, is ______.

We have, $[\sin x]=-1,0,1$

So, we have the following cases.

CASE I When $[\sin x] = -1$

In this case, we have

$x^2-2=-1⇒x=±1$

$∴x=-1$ is the solution in this case.

CASE II When $[\sin x]=0$

In this case, we have

$x^2-2=0⇒x=± \sqrt{2}$

But, $[\sin \sqrt{2}]=0$ and $[\sin (-\sqrt{2})] = −1$

$∴x=\sqrt{2}$ is the solution in this case.

CASE III When $[\sin x]=1$

In this case, we have

$x^2-2=1⇒x=±\sqrt{3}$

But, $[\sin \sqrt{3}]=0$ and $[\sin (-\sqrt{3})]=-1$. Therefore, there is no solution in this case.

Hence, the given equation has two solutions only, namely, $x=-1$ and $x = \sqrt{2}$.