`3-16x^2=0`
`<=>(\sqrt3)^2-(4x)^2=0`
`<=>(\sqrt3+4x)(\sqrt3-4x)=0`
`<=> [(\sqrt3=-4x),(\sqrt3=4x):}`
`<=> [(x=-\sqrt3/4),(x=\sqrt3/4):}`
Vậy `S={\pm \sqrt3/4}`.
Ta có: \(3-16x^2=0\)
\(\Leftrightarrow16x^2=3\)
\(\Leftrightarrow x^2=\dfrac{3}{16}\)
hay \(x\in\left\{\dfrac{\sqrt{3}}{4};-\dfrac{\sqrt{3}}{4}\right\}\)