$9x^4+3=4(2x^2+1)$
$⇔9x^4-8x^2-1=0$
$⇔(9x^4-9x^2)+(x^2-1)=0$
$⇔(x+1)(x-1)(9x^2+1)=0$
Vì $9x^2≥0∀x$ nên $9x^2+1>0∀x$
$⇒(x+1)(x-1)=0$
$⇒\left[ \begin{array}{l}x+1=0\\x-1=0\end{array} \right.⇔\left[ \begin{array}{l}x=-1\\x=1\end{array} \right.$
Vậy $S=\{-1;1\}$.