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. 

Let $$k=\frac{2-\cos x}{\sin x}\Rightarrow 2-\cos x=k\sin x$$
So we have  $$k\sin x+\cos x=2$$
Using $$|a\sin x+b\cos x|\leq \sqrt{a^2+b^2}$$
So we have $$|k\sin x+\cos x|\leq \sqrt{k^2+1}$$
$$2\leq \sqrt{k^2+1}\Rightarrow k^2+1\geq 4\Rightarrow k\geq \sqrt{3}.$$
