# Sketching graph of trigonometric function

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.

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You're right, the given answer is wrong. Since $\sin x \ge 0$ for $x\in[0, \pi]$, we have $y\le1$ in this interval.
The given answer, $x \in (\frac{\pi }{6},\frac{{5\pi }}{6})$, corresponds to the inequality $$\frac{1}{{1 - \sin (x)}} > 2,$$ where for $x=\pi/2$ the left-hand side is interpreted as $+\infty$ ($= \lim _{x \to \pi /2} \frac{1}{{1 - \sin (x)}}$). Accordingly, you should sketch the graph of $$y = \frac{1}{{1 - \sin (x)}}$$ for $x \in [0,\pi]$.
Anyway, here's a hint for the modified exercise: consider the inequality $\sin(x) > 1/2$. – Shai Covo Jun 30 '11 at 11:40