\(2{\log _4}\left( {3{x^2} + 2} \right) \ge {\log _2}\left( {2{x^2} + 5x + 2} \right)\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\log _2}\left( {3{x^2} + 2} \right) \ge {\log _2}\left( {2{x^2} + 5x + 2} \right)\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{3{x^2} + 2 \ge 2{x^2} + 5x + 2,}\\{2{x^2} + 5x + 2 > 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{{x^2}-5x \ge 0,\,\,\,\,\,\,\,\,\,\,\,}\\{2{x^2} + 5x + 2 > 0}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow } \right.} \right.\)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x\, \in \,\left( {-\infty ;0} \right] \cup \left[ {5;\infty } \right),\,\,\,\,\,\,\,\,\,}\\{x\, \in \,\left( {-\infty ;-2} \right) \cup \left( {-\dfrac{1}{2};\infty } \right)}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,x\, \in \,\left( {-\infty ;-2} \right) \cup \left( {-\dfrac{1}{2};0} \right] \cup \left[ {5;\infty } \right).} \right.\)
Ответ: \(\left( {-\infty ;-2} \right) \cup \left( {-\dfrac{1}{2};0} \right] \cup \left[ {5;\infty } \right).\)