Finding number of asymptotes in $[ 0 , 2\pi]$ interval Consider $f(x) = \frac{2\tan 2x + 1}{2\cos x + 1}$ . Now find number of asymptotes in $[ 0 , 2\pi]$ interval . I know we can simplify it to $\frac{2\sin 2x + 1}{\cos 2x (2\cos x + 1)}$ and then solve $\cos 2x (2\cos x + 1) = 0$ but is there any way for doing it straightly without simplifying ? Also I'm looking for an instruction in order to finding number of asymptotes . 

$\cos2x\neq0$ gives asymptotes $x=\frac{\pi}{4}$,  $x=\frac{3\pi}{4}$, $x=\frac{5\pi}{4}$ and $x=\frac{7\pi}{4}$.
$\cos{x}\neq-\frac{1}{2}$ gives asymptotes $x=\frac{2\pi}{3}$ and  $x=\frac{4\pi}{3}$.
Id est, we have six asymptotes.
