How to solve $\cos^2x - \sin^2x -1 = 0$? The answers from the back of the book says the answers are $\pi/2$ and $3\pi/2$. I thought that you can use the double angle formula in this problem, but it doesn't seem right? 

As you suggest, since $\cos 2x = \cos^2 x - \sin^2 x$, this is equivalent to the equation $\cos 2x = 1$, which has solutions when $2x$ is an integer multiple of $2\pi$, so the solutions are
$$x = \dots, -2\pi, -\pi, 0, \pi, 2\pi, \dots$$
So the book has the wrong answer.
