Answer
$\left( -\infty,-\dfrac{4}{3} \right)\cup[4,\infty)$
Work Step by Step
Since for any $a\gt0$, $|x|\ge a$ implies $x\ge a$ OR $x\le -a$, then the given inequality, $
\left| \dfrac{3}{4}x-1 \right|\ge2
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{3}{4}x-1\ge2
\\\\
\dfrac{3}{4}x\ge2+1
\\\\
\dfrac{3}{4}x\ge3
\\\\
\dfrac{4}{3}\cdot\dfrac{3}{4}x\ge3\cdot\dfrac{4}{3}
\\\\
x\ge4
,\\\\\text{OR}\\\\
\dfrac{3}{4}x-1\le-2
\\\\
\dfrac{3}{4}x\le-2+1
\\\\
\dfrac{3}{4}x\le-1
\\\\
\dfrac{4}{3}\cdot\dfrac{3}{4}x\le-1\cdot\dfrac{4}{3}
\\\\
x\le-\dfrac{4}{3}
.\end{array}
Hence, the solution set is $
\left( -\infty,-\dfrac{4}{3} \right)\cup[4,\infty)
.$
See graph below.