Решите уравнение: 3x^4-5x^2+2=0.

\[3x^{4} - 5x^{2} + 2 = 0\]

\[x^{2} = y \geq 0:\]

\[3y^{2} - 5y + 2 = 0\]

\[D = 25 - 24 = 1\]

\[y_{1} = \frac{5 + 1}{6} = 1;\ \]

\[y_{2} = \frac{5 - 1}{6} = \frac{4}{6} = \frac{2}{3}.\]

\[1)\ x^{2} = 1\]

\[x = \pm 1.\]

\[2)\ x^{2} = \frac{2}{3}\]

\[x = \pm \sqrt{\frac{2}{3}} = \pm \frac{\sqrt{6}}{3}.\]

\[Ответ:x = \pm \frac{\sqrt{6}}{3};\ \ x = \pm 1.\]