Solving a quadratic trigonometric inequality Say I have the following inequality:
$$\cos^2x-\cos{x}>0$$
My books says the inequality is verified for 
$$\cos{x}<0\space\vee\cos{x}>1$$
How does one end up to that conclusion?
I know that the inequality above is equal to
$$\cos{x}(\cos{x}-1)>0 \rightarrow \cos{x} > 0 \vee \cos{x}>1$$
So what am I doing wrong?

$$
\begin{align}
\cos x(\cos x-1)\gt0
&
\implies
\big( \cos x\gt 0\land \cos x-1\gt 0 \big)\lor\big(\cos x\lt 0\land \cos x-1\lt 0\big)
\\
&\implies 
\big( \cos x\gt 0\land \cos x\gt 1 \big)\lor\big(\cos x\lt 0\land \cos x\lt 1\big)
\end{align}
$$
The condition $\cos x\gt 1$ never holds since $\cos x$ always oscillates between $-1$ and $1$. Hence the only remaining possibility is $\cos x\lt0\land\cos x\lt 1$.
