Official solution
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).