Answer
$\left( -\infty, \dfrac{2}{3}\right] \cup\left[ 2,\infty \right)$
Work Step by Step
Since for any $a\gt0$, $|x|\gt a$ implies $x \ge a$ or $x\le -a$ then the solution to the given inequality, $
\left| 3x-4 \right|\ge 2
,$ is
\begin{array}{l}\require{cancel}
3x-4\ge 2
\\\\
3x\ge 2+4
\\\\
3x\ge 6
\\\\
x\ge \dfrac{6}{3}
\\\\
x\ge 2
,\\\\\text{OR}\\\\
3x-4\le -2
\\\\
3x\le -2+4
\\\\
3x\le 2
\\\\
x\le \dfrac{2}{3}
.\end{array}
Hence, the solution set is the interval $
\left( -\infty, \dfrac{2}{3}\right] \cup\left[ 2,\infty \right)
.$
