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