Minimum value of $\frac{2-\cos(x)}{\sin(x)}$ without differentiation I have to find the minimum value of the expression 
$$\frac{2 - \cos x}{ \sin x}$$
Also $x$ lies between $0$ to $\pi$. One way is to find the minima using differentiation. But it is not taught in my grade so my teacher asked me to do it without differentiation.
Here's what I did 

Let$$\frac{2 - \cos x }{ \sin x} = y~,$$
so that $$ (2-y \sin x)^2 = 1 - \sin^2(x) \\
\implies \sin^2(x) \cdot (y^2 +1) - 4 y \sin x +3=0$$
Now I am struck. I tried using Discriminant $\ge{0}$ but no use as our variable $\sin x$  lies between $0$and $1$. Please help. 

You can use AM-GM as follows for $x \in (0,\pi)$:
$$\frac{2-\cos(x)}{\sin(x)}= \frac{2-(\cos^2(x/2) - \sin^2(x/2))}{2\sin(x/2)\cos(x/2)}$$
$$= \frac{\cos^2(x/2) +3\sin^2(x/2)}{2\sin(x/2)\cos(x/2)}\stackrel{AM-GM}{\geq} \frac{\sqrt{\cos^2(x/2) \cdot 3\sin^2(x/2)}}{\sin(x/2)\cos(x/2)} = \sqrt{3}$$
Additional note after comment:
Note that according to AM-GM you have equality if and only if
$$\cos^2(x/2) = 3\sin^2(x/2) \Leftrightarrow \tan^2(x/2) = \frac{1}{3} \stackrel{x \in (0,\pi)}{\Leftrightarrow}x = \frac{\pi}{3}$$
