I need an explanation of the end of the problem All I need is to know how they get to pi / 6 (what's in the circle)
note:sen x =sin x
note: it is a single problem that uses trigonometry along with its development

[problem][2]
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problem
 A: For $x\in (0,\pi /2)$: Let $f(x)=(\sin x)/x.$ Then $f'(x)=(x\cos x -\sin x)/x^2=(x-\tan x)/(x^2\cos x)<0.$
So $f$ is strictly decreasing so there is at most one $x\in (0,\pi/2)$ such that $f(x)=3/\pi.$
And there is at least one, as $f(\pi /6)=3/\pi.$
So $[x\in (0,\pi/2)\land \sin x=3x/\pi]\iff [x\in (0,\pi/2)\land f(x)=3/\pi]\iff  [x=\pi/6].$
A: The book jumped to a quick conclusion in solution of the equation:
$$\dfrac{\sin \alpha}{\alpha}=\dfrac{3}{\pi}$$
He perhaps expects you to find or verify this by a process of trial and error. 
You are right, that you got this query.. that it cannot be solved using pure trigonometry alone. Had it been some other value in the given problem it would have entailed solving a transcendental equation by numerical methods that may not have been in the scope at the present stage of learning.
A: If 
$$\sin \alpha = \dfrac{3 \alpha}{\pi} \tag{1},$$
it is true that $\alpha = \dfrac{\pi}{6}$ is a solution (the left hand side and the right hand side of (1) are equal to $ \dfrac12$).
But there are two other solutions : $\alpha = 0$ and $\alpha = \dfrac{\pi}{6}$ as one can see on a graphical representation with the (very close) curves of $y=\sin \alpha$ and $y=\dfrac{3}{\pi} \alpha$ (which is a straight line with a slope slightly less than $1$)
Your textbook should have given the two other solutions and say that they are not "physically" meaningful.
