Ans:- D Hint:- We have [sin x] = -1 , 0 , 1 So, we have the following cases Case – I :- when [sin x] = -1, In this case , we have x2 - 2 = -1 ⇒ x ± 1 ∴ x = - 1 is the solution in this case Case – II :- when [sin x] = 0 , In this case, we have x2 - 2 = 0 ⇒ x ± √2 But , [sin √2] = 0 and [sin(−√2)] = - 1 ∴ x = √2 is the solution in this case Case – III :- when [sin x] = 1, In this case, we have x2 - 2 = 1 ⇒ x = ± √3 But , [sin √3] = 0 and [sin (−√3)] = - 1 , Therefore, there is no solution in this case. Hence, the given equation has two solutions only, namely, x = - 1 and x = √2