The number of solutions of the equation

$${\sin ^{ - 1}}\left[ {{x^2} + {1 \over 3}} \right] + {\cos ^{ - 1}}\left[ {{x^2} - {2 \over 3}} \right] = {x^2}$$, for x$\in$[$-$1, 1], and [x] denotes the greatest integer less than or equal to x, is :

There are three cases possible for $x \in [ - 1,1]$<br><br>Case I : $x \in \left[ { - 1, - \sqrt {{2 \over 3}} } \right)$<br><br>$\therefore$ ${\sin ^{ - 1}}(1) + {\cos ^{ - 1}}(0) = {x^2}$<br><br>$\Rightarrow {x^2} = {\pi \over 2} + {\pi \over 2} = \pi$<br><br>$\Rightarrow x = \pm \sqrt \pi$ $\to$ (Reject)<br><br>Case II : $x \in \left( { - \sqrt {{2 \over 3}} ,\sqrt {{2 \over 3}} } \right)$<br><br>$\therefore$ ${\sin ^{ - 1}}(0) + {\cos ^{ - 1}}( - 1) = {x^2}$<br><br>$\Rightarrow 0 + \pi = {x^2}$<br><br>$\Rightarrow x = \pm \sqrt x$ $\to$ (Reject)<br><br>Case III : $x \in \left( {\sqrt {{2 \over 3}} ,1} \right)$<br><br>$\therefore$ ${\sin ^{ - 1}}(0) + {\cos ^{ - 1}}(0) = {x^2}$<br><br>$\Rightarrow {x^2} - \pi \Rightarrow x - \pm \sqrt x$ (Reject)<br><br>$\therefore$ No solution. There, the correct answer is (1).