To solve the equation \( \cos^2 2x = \cos^2 \left(x + \frac{\pi}{6}\right) \), we can proceed as follows:

1. **Using the double-angle identity for cosine:**
\[ \cos^2 2x = \frac{1 + \cos 4x}{2} \]

2. **Expanding \( \cos^2 \left(x + \frac{\pi}{6}\right) \):**
\[ \cos^2 \left(x + \frac{\pi}{6}\right) = \frac{1 + \cos \left(2x + \frac{\pi}{3}\right)}{2} \]

3. **Setting the equations equal to each other:**
\[ \frac{1 + \cos 4x}{2} = \frac{1 + \cos \left(2x + \frac{\pi}{3}\right)}{2} \]

4. **Eliminate the denominators:**
\[ 1 + \cos 4x = 1 + \cos \left(2x + \frac{\pi}{3}\right) \]

5. **Simplify the equation:**
\[ \cos 4x = \cos \left(2x + \frac{\pi}{3}\right) \]

6. **Apply the cosine identity:**
\[ 4x = 2x + \frac{\pi}{3} + 2k\pi \quad \text{or} \quad 4x = - (2x + \frac{\pi}{3}) + 2k\pi \]

Where \( k \) is any integer.

7. **Solve for \( x \):**

- From \( 4x = 2x + \frac{\pi}{3} + 2k\pi \):
\[ 2x = \frac{\pi}{3} + 2k\pi \]
\[ x = \frac{\pi}{6} + k\pi \]

- From \( 4x = - (2x + \frac{\pi}{3}) + 2k\pi \):
\[ 4x = -2x - \frac{\pi}{3} + 2k\pi \]
\[ 6x = -\frac{\pi}{3} + 2k\pi \]
\[ x = -\frac{\pi}{18} + \frac{k\pi}{3} \]

Therefore, the solutions for \( x \) are:

\[ x = \frac{\pi}{6} + k\pi \quad \text{or} \quad x = -\frac{\pi}{18} + \frac{k\pi}{3} \]

where \( k \) is any integer. These represent the general solutions to the equation \( \cos^2 2x = \cos^2 \left(x + \frac{\pi}{6}\right) \).