The number of solutions of the equation ^ - 1 [ x ^ 2 + (1)/(3)] +...

Given equation is

$\sin^{- 1}$\left[{x}^{2} + \dfrac{1}{3}\right] + \cos^{- 1}$\left[{x}^{2} - \dfrac{2}{3}\right] = {x}^{2}$

Now, $\sin^{- 1}\left[{x}^{2} + \dfrac{1}{3}\right]$ is defined if $- 1 \le \left[{x}^{2} + \dfrac{1}{3}\right] \le 1$

$⇒ - 1 \le {x}^{2} + \dfrac{1}{3} < 2$

$⇒ \dfrac{- 4}{3} \le {x}^{2} < \dfrac{5}{3}$

$⇒ 0 \le {x}^{2} < \dfrac{5}{3} . . . \left(1\right)$

Also, and $\cos^{- 1}\left[{x}^{2} - \dfrac{2}{3}\right]$ is defined if $- 1 \le \left[{x}^{2} - \dfrac{2}{3}\right] \le 1$

$⇒ - 1 \le {x}^{2} - \dfrac{2}{3} < 2$

$⇒ \dfrac{- 1}{3} \le {x}^{2} < \dfrac{8}{3}$

$⇒ 0 \le {x}^{2} < \dfrac{8}{3} . . . \left(2\right)$

So, from $( 1 )$ and $( 2 )$ we can conclude $0 \le {x}^{2} < \dfrac{5}{3}$

Case -$I:$ If $0 \le {x}^{2} < \dfrac{2}{3}$

$⇒ \sin^{- 1} ( 0 ) + \cos^{- 1} ( - 1 ) = {x}^{2}$

$⇒ 0 + \pi = {x}^{2}$

$⇒ {x}^{2} = \pi$ but $\pi \notin \left[0, \dfrac{2}{3}\right)$

$⇒$ No value of $' x '$

Case - $II:$ If $\dfrac{2}{3} \le {x}^{2} < \dfrac{5}{3}$

$⇒ \sin^{- 1} ( 1 ) + \cos^{- 1} ( 0 ) = {x}^{2}$

$⇒ \dfrac{\pi}{2} + \dfrac{\pi}{2} = {x}^{2}$

$⇒ {x}^{2} = \pi$ but $\pi \notin \left[\dfrac{2}{3}, \dfrac{5}{3}\right)$

$⇒$ No value of $' x '$

So, number of solutions of the equation is zero.