To solve the equation \( \sin(2x) + \cos(4x) = 0 \), we can use some trigonometric identities.

First, rewrite \( \cos(4x) \) using the double angle formula:

\[
\cos(4x) = \cos^2(2x) - \sin^2(2x) = 2\cos^2(2x) - 1
\]

Thus, the equation becomes:

\[
\sin(2x) + 2\cos^2(2x) - 1 = 0
\]

We also know that:

\[
\sin(2x) = 2\sin(x)\cos(x)
\]

Let's substitute \( y = \sin(2x) \):

\[
y + 2\cos^2(2x) - 1 = 0
\]

Now, recognize that \( cos^2(2x) = 1 - \sin^2(2x) \).

We can now rewrite the original equation for further exploration, using the relationship between \( \sin(2x) \) and \( \cos(4x) \).

Instead, we could plot or numerically find solutions by considering:

1. Rearranging the equation:
\[
\cos(4x) = -\sin(2x)
\]

Since the trigonometric functions are periodic, we can set:
- The solutions might occur where both sides have the same values by utilizing angles of sine and cosine.

2. Solve \( \sin(2x) = -\cos(4x) \):
- For example, utilizing \( \sin(2x) - \cos(4x) = 0 \).

To solve \( \sin(2x) + 1 = 0 \) and \( \cos(4x) = -1 \):

### 1. Solving \( \sin(2x) = -1 \):
\[
2x = \frac{3\pi}{2} + 2k\pi \quad \text{for } k \in \mathbb{Z}
\]
\[
x = \frac{3\pi}{4} + k\pi
\]

### 2. Solving \( \cos(4x) = 0 \):
\[
4x = \frac{\pi}{2} + n\pi \quad \text{for } n \in \mathbb{Z}
\]
\[
x = \frac{\pi}{8} + \frac{n\pi}{4}
\]

### Final Solution
Thus, the general solution for \( x \) is:
1. \( x = \frac{3\pi}{4} + k\pi \) for \( k \in \mathbb{Z} \).
2. \( x = \frac{\pi}{8} + \frac{n\pi}{4} \) for \( n \in \mathbb{Z} \).

These will give the complete set of solutions for the equation \( \sin(2x) + \cos(4x) = 0 \).