Answer
$\{\pm\frac{\sqrt 2}{2},\pm \frac{\sqrt {6}}{2}\}$
Work Step by Step
Step 1. Let $u=x^2$, we have $4u^2-8u+3=0$
Step 2. Factor and solve: $(2u-1)(2u-3)=0$ thus $u= \frac{1}{2}, \frac{3}{2}$
Step 3. For $u= \frac{1}{2}$, we have $x^2= \frac{1}{2}$ and $x=\pm \frac{\sqrt 2}{2}$
Step 4. For $u=\frac{3}{2}$, we have $x^2=\frac{3}{2}$ and $x=\pm \frac{\sqrt {6}}{2}$
Step 5. Solution set $\{\pm\frac{\sqrt 2}{2},\pm \frac{\sqrt {6}}{2}\}$
