The function, sign of which we are asked to study, is:
$$f(x) = 2\sin(x)\cos(x) - 1,$$
or, simplified:
$$f(x) = \sin(2x)-1,$$
without using its graph to do so, since the exercise itself asks, in the end, to plot the function's behaviour in a approximate graph.
If I were to study its sign how I would usually do for more complex functions, I would say:
$$\sin(2x) \geq 1,$$
which never happens, excluding the case in which $\sin(2x) = 1$ .
At this point, I would likely say that the function is never greater than 1, and is, hence, always negative. Am I doing anything wrong here?
... Figured out as I was rereading my own question:
-1is the function's middle line, which shifts the whole graph down by-1. The inequality I started with has no solutions because, indeed, the function never evaluates to aygreater than zero. I was greatly anxious, and as I just can't handle these kinds of situations, mistakes like this I guess may happen! Didn't delete the question, because hey, it might eventually help anybody who'd find him/herself in the same, apparently stupid, situation. :) Cheers!