For $\sqrt{x\left(x+1\right)}$ to be defined $x\left(x+1\right)\ge 0 ...\left(i\right)$

Also, for ${\sin }^{-1}x,$ we know that $x\in \left[-1, 1\right]$

 $\Rightarrow x^2+x+1\le 1$

$\Rightarrow x^2+x\le 0 ...\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right),$ we get

$x\left(x+1\right)=0$

$\Rightarrow x=0,-1$

When $x=0,$

$L.H.S.={\tan }^{-1}0+{\sin }^{-1}1=\frac{\pi }{2}$

And, when $x=-1,$

$L.H.S.={\tan }^{-1}0+{\sin }^{-1}\sqrt{1-1+1}$

$=0+{\sin }^{-1}\left(1\right)=\frac{\pi }{2}$

Thus, the number of solutions are $2.$