Cos^2x - sin^2x = -1

To solve the equation \( \cos^2 x - \sin^2 x = -1 \), we can use the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \).

Starting from the given equation, we can rewrite \( \cos^2 x - \sin^2 x \) as:

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

This can be rewritten by rearranging terms:

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

Now, notice that we can express \( \cos^2 x \) in terms of \( \sin^2 x \) using the Pythagorean identity. Let’s denote \( \sin^2 x = y \). Thus, \( \cos^2 x = 1 - y \).

Now substituting this into the equation gives:

\[
(1 - y) - y + 1 = 0
\]

Simplifying this:

\[
1 - y - y + 1 = 0 \implies 2 - 2y = 0 \implies 2y = 2 \implies y = 1
\]

Since we defined \( y = \sin^2 x \), we have \( \sin^2 x = 1 \). Taking the square root gives:

\[
\sin x = \pm 1
\]

The sine function equals 1 at:

\[
x = \frac{\pi}{2} + 2k\pi \quad \text{(for } k \in \mathbb{Z}\text{)}
\]

The sine function equals -1 at:

\[
x = \frac{3\pi}{2} + 2k\pi \quad \text{(for } k \in \mathbb{Z}\text{)}
\]

Therefore, the complete set of solutions for the equation \( \cos^2 x - \sin^2 x = -1 \) is:

\[
x = \frac{\pi}{2} + 2k\pi \quad \text{and} \quad x = \frac{3\pi}{2} + 2k\pi \quad \text{(for } k \in \mathbb{Z}\text{)}
\]