We need to solve the equation:

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

We can start by using the identity for the difference of squares:

\[
a^2 - b^2 = (a-b)(a+b)
\]

Let \( a = \cos^2 x \) and \( b = \sin^2 x \), then:

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

Since \( \cos^2 x + \sin^2 x = 1 \), the equation simplifies to:

\[
\cos^4 x - \sin^4 x = \cos^2 x - \sin^2 x
\]

Now we set this equal to the right side of the original equation:

\[
\cos^2 x - \sin^2 x = 2\cos^2 x - 1
\]

Rearranging gives:

\[
-\sin^2 x = \cos^2 x - 1
\]

Since \( \cos^2 x = 1 - \sin^2 x \), we can substitute \( \cos^2 x \):

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

Simplifying this results in:

\[
-\sin^2 x = -\sin^2 x
\]

This means the equation holds true for all \( x \).

Thus, the solution to \( \cos^4 x - \sin^4 x = 2\cos^2 x - 1 \) is all real values of \( x \).