To solve \(|x| < |5 + 2x|\), consider the critical points where \(x = 0\) and \(5 + 2x = 0\).

First, solve \(5 + 2x = 0\):

\(2x = -5\)

\(x = -\frac{5}{2}\)

Now, consider the intervals determined by the critical points \(x = 0\) and \(x = -\frac{5}{2}\).

1. For \(x < -5\), both \(x\) and \(5 + 2x\) are negative, so \(|x| = -x\) and \(|5 + 2x| = -(5 + 2x)\).

2. For \(-5 < x < -\frac{5}{3}\), \(x\) is negative and \(5 + 2x\) is positive, so \(|x| = -x\) and \(|5 + 2x| = 5 + 2x\).

3. For \(x > -\frac{5}{3}\), both \(x\) and \(5 + 2x\) are positive, so \(|x| = x\) and \(|5 + 2x| = 5 + 2x\).

Check each interval:

- For \(x < -5\), \(-x < -(5 + 2x)\) simplifies to \(x < -5\).

- For \(-5 < x < -\frac{5}{3}\), \(-x < 5 + 2x\) simplifies to \(x > -\frac{5}{3}\).

- For \(x > -\frac{5}{3}\), \(x < 5 + 2x\) simplifies to \(x < -5\), which is not possible.

Thus, the solution is \(x < -5\) or \(x > -\frac{5}{3}\).