I am a little confused with graphing this trigonometric function.
Sketch the graph of $y = \dfrac{1}{1+\sin{(x)}}$ for $0 \le x \le \pi$. Find the set of values of x, lying in the interval $0 \le x \le \pi$ for which $\dfrac{1}{1+\sin{(x)}} > 2$.
I was able to graph the function with vertical asymptotes at $x = -\dfrac{\pi}{2} \pm 2\pi$ and horizontal asymptote at $y = 0$. The graph is like a repeating U pattern.
My question is with regards to finding the set of values of x. My estimate is that the line $y = 2$ does not cut it in the interval $[0, \pi]$. The intersection is at $x = -\dfrac{\pi}{6}$ which isn't in this interval.
However the given answer is $x \in \left(\dfrac{\pi}{6}, \dfrac{5\pi}{6}\right)$. I am unclear how to get this. Can you guys clarify any mistakes I may have made? Thanks for your help.