The domain is $x \in [-1,1]$. We recognize that LHS is strictly decreasing, while RHS is strictly increasing from $x=-1$ to $x=0$ and decreasing from $x=0$ to $x=1$. The LHS has a value of $2$ when $x=-1$ and a value of $1$ when $x=0$, while RHS is $1$ when $x=-1$ and is $2$ when $x=0$. This means that the two sides meet only once. The equation never holds when $x>0$ because $LHS < 1$ while $RHS \geq 1$ in this case. As for what the solution is, I've tried substituting in $\sqrt{1-x^2}=\sin (\arccos x)$ and now all that's needed is to solve the equation $\frac{2}{\pi} y = \sin y + 1$ for $0 \leq y \leq \pi$.