To solve the equation \((9\sqrt{x} - 4)(x^2 + 4) = 0\), we can set each factor to zero.

1. **Setting the first factor to zero**:
\[
9\sqrt{x} - 4 = 0
\]
Add 4 to both sides:
\[
9\sqrt{x} = 4
\]
Divide by 9:
\[
\sqrt{x} = \frac{4}{9}
\]
Now, square both sides to solve for \(x\):
\[
x = \left(\frac{4}{9}\right)^2 = \frac{16}{81}
\]

2. **Setting the second factor to zero**:
\[
x^2 + 4 = 0
\]
Subtract 4 from both sides:
\[
x^2 = -4
\]
This equation has no real solutions since the square of a real number cannot be negative. However, if we consider complex numbers, we can write:
\[
x = \pm 2i
\]

In summary, the solutions to the given equation are:
- \(x = \frac{16}{81}\) (real solution)
- \(x = 2i\) (complex solution)
- \(x = -2i\) (complex solution)