Answer
$\left( -\infty, -\dfrac{22}{15} \right]\cup\left[ \dfrac{6}{5},\infty \right)$
Work Step by Step
Since for any $a\gt0$, $|x|\ge a$ implies $x\ge a\text{ or } x\le-a$, then the expression, $
\left| 3x+\dfrac{2}{5} \right|\ge4
,$ is equivalent to
\begin{array}{l}\require{cancel}
3x+\dfrac{2}{5}\ge4
\\\\
3x\ge4-\dfrac{2}{5}
\\\\
3x\ge\dfrac{20}{5}-\dfrac{2}{5}
\\\\
3x\ge\dfrac{18}{5}
\\\\
x\ge\dfrac{\dfrac{18}{5}}{3}
\\\\
x\ge\dfrac{18}{15}
\\\\
x\ge\dfrac{6}{5}
,\\\\\text{ OR }\\\\
3x+\dfrac{2}{5}\le-4
\\\\
3x\le-4-\dfrac{2}{5}
\\\\
3x\le-\dfrac{20}{5}-\dfrac{2}{5}
\\\\
3x\le-\dfrac{22}{5}
\\\\
x\le-\dfrac{\dfrac{22}{5}}{3}
\\\\
x\le-\dfrac{22}{15}
.\end{array}
Hence, the solution set is $
\left( -\infty, -\dfrac{22}{15} \right]\cup\left[ \dfrac{6}{5},\infty \right)
.$
