đặt $\sqrt{1-x^2}=t,t\geq 0$$\Rightarrow \begin{cases}x^2+t^2=1\\ 729x^4+8t=36 \end{cases} $
$\Rightarrow 36(x^2+t^2)=729x^4+8t \Leftrightarrow 729x^4-36x^2-36t^2+8t=0$
$\triangle ' = (162t-18)^2$
$\Rightarrow x^2=\frac{2}{9}t \vee x^2=\frac{4}{81}-\frac{2}{9}t $