Why this system have one solution Let $b\in (1,2),x\in (0,\frac{\pi}{2})$,if such 
$$\begin{cases}
2b^2+b-4=2\sqrt{4-b^2}\cos{x}\\
2b^2-4=2b\cos{(x+\frac{\pi}{18})}-2\sqrt{4-b^2}\cos{\frac{5\pi}{18}}
\end{cases}$$
show that:$$x=\dfrac{\pi}{6}$$ 
Here is what I already got.
First of all, one should notice equation $x=\dfrac{\pi}{6}$,
$$2b^2+b-4=\sqrt{12-3b^2}$$
then $b$ such
$$b^3-3b+1=0$$
But this kind of proof does not fit my appetite as it not only involves some additional theorem but also not very nice as the simple nice form of the question of itself.

Hint $cos50\approx 0.6$ so create $cos(x+π/18)$ in terms of b from second equation and create $cosx$ in terms of b from first equation creating them and simplifying you get a huge equation! Which is $$8b^4+4b^3-18b^2+4b^2\sqrt{4-b^2}+6.6b\sqrt{4-b^2}-26.4\sqrt{4-b^2}+4=0$$ solving it you get roots but only $1.15023$ satisfies this as  $b$ belongs to $(1,2)$ so its $1.42$ then approximately due to numerical approximation you get $cos(x)=0.62$ thus $x=π/6$
