I don't understand how the solution of $\cot x dy = (2-y)dx; y(x) \to -1, x\to0$ doesn't contain absolute value signs

Question

So I have this system $\cot x dy = (2-y)dx; y(x) \to -1, x\to0$. I modify the equation to $(2-y)^{-1} dy = \tan x dx$, then integrate both sides and eventually I get $-\ln |2-y| = ln|1/\ cos x| + C$ and $|\ln |2-y| = \ln |\cos x| + C$ finally $|2-y| = C |\cos x|$, $C = const$. Throughout the solution some cases get lost when dividing both sides on functions (like the case $\cot x = 0$ needs to be checked, or the function $y=2$). Assuming those have been checked too, what I don't get is how my textbook gets rid of the absolute value signs. What I mean is that the answer is shown as $y = 2 + C \cos x$ (and for the system by passing to limit we'll get $y = 2 + 3 \cos x$). I don't understand how the book got rid of the absolute value signs.