The number of solutions of the equation ${\sin }^{-1}\left[{\mathrm{x}}^2+\frac{1}{3}\right]+{\cos }^{-1}\left[{\mathrm{x}}^2-\frac{2}{3}\right]={\mathrm{x}}^2,\text{ for }\mathrm{x}\in [-1,1]$ and [x ] denotes the greatest integer less than or equal to x , is

$$\begin{gathered} \mathrm{Given} \mathrm{equation} \mathrm{is} \\ {\sin }^{-1}\left[x^2+\frac{1}{3}\right]+{\cos }^{-1}\left[x^2-\frac{2}{3}\right]=x^2 \end{gathered}$$

$$\begin{gathered} {\sin }^{-1}\left[x^2+\frac{1}{3}\right] \mathrm{is}\mathrm{defined} i\mathrm{f} \\ -1\le x^2+\frac{1}{3}<2\Rightarrow \frac{-4}{3}\le x^2<\frac{5}{3}\Rightarrow 0\le x^2<\frac{5}{3} \end{gathered}$$

$$\begin{gathered} {\cos }^{-1}\left[x^2-\frac{2}{3}\right]\text{ is defined if } \\ -1\le {\mathrm{x}}^2-\frac{2}{3}<2\Rightarrow \frac{-1}{3}\le {\mathrm{x}}^2<\frac{8}{3}\Rightarrow 0\le {\mathrm{x}}^2<\frac{8}{3} \\ \Rightarrow 0\le {\mathrm{x}}^2<\frac{5}{3} \end{gathered}$$

$\begin{array}{l}\text{ Case }-\mathbf{I}\text{ if }0\le x^2<\frac{2}{3}\\ {\sin }^{-1}\left(0\right)+{\cos }^{-1}(-1)=x^2\\ \Rightarrow 0+\pi =x^2\\ \Rightarrow x^2=\pi \\ \text{ but }\pi \notin \left[0,\frac{2}{3}\right)\Rightarrow \text{ No value of ' }x\text{ ' }\end{array}$

$\begin{array}{l}\text{ Case - II if }\frac{2}{3}\le x^2<\frac{5}{3}\\ {\sin }^{-1}\left(1\right)+{\cos }^{-1}\left(0\right)=x^2\\ \Rightarrow \frac{\pi }{2}+\frac{\pi }{2}=x^2\\ \Rightarrow x^2=\pi \\ \begin{array}{r}\text{ but }\pi \notin \left(\frac{2}{3},\frac{3}{3}\right)\\ \Rightarrow \text{ No value of ' }x\text{ ' }\end{array}\end{array}$

$$Number of solutions is Zero.