Answer: \(x=3\) or \(x=-\frac12\).

Use the two possible cases for the modulus.

If \(3x-2\geq0\), then

\(3x-2=4+x.\)

So

\(2x=6,\)

and hence

\(x=3.\)

If \(3x-2\lt0\), then

\(-(3x-2)=4+x.\)

This gives

\(2-3x=4+x.\)

So

\(-4x=2,\)

and hence

\(x=-\frac12.\)

Both values satisfy the original equation, so the solutions are

\(x=3\quad\text{or}\quad x=-\frac12.\)