Prove an inequality with a $\sin$ function: $\sin(x) > \frac2\pi x$ for $0$$\forall{x\in(0,\frac{\pi}{2})}\ 
\sin(x) > \frac{2}{\pi}x  $$
I suppose that solving $ \sin x = \frac{2}{\pi}x $  is the top difficulty of this exercise, but I don't know how to think out such cases in which there is an argument on the right side of a trigonometric equation. 

Here is a simple solution.
Let $0 < x < \pi/2$ be fixed. By Mean value theorem
there exists $y \in (0, x)$ such that
$$\sin(x)-\sin(0)= \cos(y)(x-0).$$
Thus $$\frac{\sin(x)}{x}= \cos(y).$$
Similarly  there exists $z \in (x, \pi/2)$ such that
$$\sin(\pi/2)-\sin(x)= \cos(z)(\pi/2-x).$$
Thus $$\frac{1-\sin(x)}{\pi/2- x}= \cos(z).$$
As $0 < y < z< \pi/2$ and $\cos$ is a strictly decreasing function in $[0, \pi/2]$ we see that
$$ \cos (z) < \cos (y).$$
Thus $$\frac{1-\sin(x)}{\pi/2- x} < \frac{\sin(x)}{x}.$$
$$\Rightarrow \frac{1-\sin(x)}{\sin(x)} < \frac{\pi/2- x}{x}.$$
$$\Rightarrow \frac{1}{\sin(x)}- 1< \frac{\pi}{2x}- 1.$$
$$\Rightarrow \frac{2}{\pi}< \frac{\sin (x)}{x}.$$
Since $y \in (0, \pi/2)$, therefore $\cos(y) <1$. Hence it is also clear that
$$\frac{\sin(x)}{x}= \cos(y) <1.$$
Hence for any $0 < x < \pi/2$, we get
$$\frac 2 \pi<\frac{\sin(x)}{x}<1.$$
