((căn 1+x)+(căn 1-x))(2+2 căn (1- x^2)) = 8

ĐK: $-1\le x\le 1$

$(\sqrt{1+x}+\sqrt{1-x})(2+2\sqrt{1-x^2})=8\\⇔(\sqrt{1+x}+\sqrt{1-x})[(1+x)+2\sqrt{(1+x)(1-x)}+(1-x)]=8\\⇔(\sqrt{1+x}+\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})^2=8\\⇔(\sqrt{1+x}+\sqrt{1-x})^3=8\\⇔\sqrt{1+x}+\sqrt{1-x}=2\\⇔2+2\sqrt{1-x^2}=4\\⇔2\sqrt{1-x^2}=2\\⇔\sqrt{1-x^2}=1\\⇔1-x^2=1\\⇔x^2=0\\⇔x=0(TM)$

Vậy $S=\{0\}$