To solve \(|2x + 3| > 3|x + 2|\), consider the non-modular inequality:

\((2x + 3)^2 > 3^2(x + 2)^2\)

Expanding both sides gives:

\((2x + 3)^2 = 4x^2 + 12x + 9\)

\(3^2(x + 2)^2 = 9(x^2 + 4x + 4) = 9x^2 + 36x + 36\)

Set up the inequality:

\(4x^2 + 12x + 9 > 9x^2 + 36x + 36\)

Simplify to:

\(4x^2 + 12x + 9 - 9x^2 - 36x - 36 > 0\)

\(-5x^2 - 24x - 27 > 0\)

Multiply through by -1 (reversing the inequality):

\(5x^2 + 24x + 27 < 0\)

Factor the quadratic:

\((5x + 9)(x + 3) < 0\)

Find critical points by setting each factor to zero:

\(5x + 9 = 0 \Rightarrow x = -\frac{9}{5}\)

\(x + 3 = 0 \Rightarrow x = -3\)

Test intervals around the critical points to determine where the inequality holds:

For \(-3 < x < -\frac{9}{5}\), the inequality holds.

For \(x > -3\) and \(x < -\frac{9}{5}\), the inequality holds.

Thus, the solution is:

\(-3 < x < \frac{9}{5} \text{ or } x > -3 \text{ and } x < \frac{9}{5}\)