Find the signal and zeros of $\cos x - \sin x$ 
Let $f$ be a function of domain $\mathbb{R}$ defined by:
$$f(x) = \cos x - \sin x$$
Find the signal and zeros of the function.

First I tried to find the zeros of $f(x)$:
$$0 = \cos x - \sin x \Leftrightarrow \\
0 = \sqrt{1-\sin^2x}-\sin x \Leftrightarrow \\
\sin x = \sqrt{1-\sin^2x} \Leftrightarrow \\
\sin^2x=1-\sin^2x\Leftrightarrow \\
2\sin^2x=1 \Leftrightarrow \\
\sin x = \frac{1}{\sqrt{2}} \Leftrightarrow \\
x = \frac{\pi}{4}+2k\pi \lor \pi - \frac{\pi}{4}+2k\pi \Leftrightarrow \\
x = \frac{\pi}{4}+2k\pi \lor \frac{3\pi}{4}+2k\pi $$
But then when I check the value of $f(x)$ for the answers above, $\frac{\pi}{4}+2k\pi$ gives me a zero while $\frac{3\pi}{4}+2k\pi$ gives me a minimum. Why is that?
As for the signal, $\frac{\pi}{4}+2k\pi$ gives me the zero on the left side when the function becomes negative.
My book states the solution is 

Zeros:
$x = \frac{\pi}{4}+k\pi, k\in \mathbb{R}$
$f$ is negative in the intervals 
$$]\frac{\pi}{4}+k2\pi,\frac{5\pi}{4}+k2\pi[,k\in\mathbb{Z};$$
$f$ is positive in the intervals
$$]-\frac{3\pi}{4}+k2\pi,\frac{\pi}{4}+k2\pi[,k\in\mathbb{Z}$$

How do I solve this?

$f(x)=\cos x−\sin x = \sqrt {2} \cos (x+\frac \pi 4)$
more generally
$a\cos x+b\sin x = \sqrt {a^2 + b^2} \cos (x - \tan^{-1}\frac ba)$ 
