Answer: \(x\leqslant -1\) or \(x\geqslant \frac15\).

For a modulus inequality of the form \(|A|\geqslant3\), the expression \(A\) must be at least \(3\) or at most \(-3\).

Here

\(|5x+2|\geqslant3.\)

So either

\(5x+2\geqslant3\)

or

\(5x+2\leqslant-3.\)

Solving the first inequality gives

\(5x\geqslant1,\)

so

\(x\geqslant\frac15.\)

Solving the second inequality gives

\(5x\leqslant-5,\)

so

\(x\leqslant-1.\)

Therefore the solution is

\(x\leqslant-1\quad\text{or}\quad x\geqslant\frac15.\)