Answer
$[0,\frac{4}{3}]$
See graph.
Work Step by Step
Step 1. Remove the absolute value sign and transform the original inequality:
$2+|2-3x|\le4\Longrightarrow |3x-2|\le2 \Longrightarrow -2\le 3x-2\le2 \Longrightarrow 0\le 3x\le4 \Longrightarrow 0\le x\le\frac{4}{3}$, using interval form $[0,\frac{4}{3}]$
Step 2. See graph.
