<s>[INST] 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.
 [/INST] In the spirit as Ian Miller's answer, the first equation leads to $$\cos(x)=\frac{2 b^2+b-4}{2 \sqrt{4-b^2}}$$ In the second equation, replace $$\cos{(x+\frac{\pi}{18})}=\cos \left(\frac{\pi }{18}\right) \cos (x)-\sin \left(\frac{\pi }{18}\right) \sin (x)$$ which allows to extract $\sin(x)$ given by $$-\frac{\csc \left(\frac{\pi }{18}\right) \left(-4 \sqrt{4-b^2}-b \left(-2
   \sqrt{4-b^2} b+\left(2 b^2+b-4\right) \cos \left(\frac{\pi }{18}\right)+2 b \sin
   \left(\frac{2 \pi }{9}\right)\right)+8 \sin \left(\frac{2 \pi
   }{9}\right)\right)}{2 b \sqrt{4-b^2}}$$ (nice monster !). Now (do not try to write it !) consider $$f(b)=\sin^2(x)+\cos^2(x)-1=0$$ and plot the function. You will notice that, in the range $1\leq b \leq 2$, there are two roots to the equation. Solving numerically $$b_1\approx 1.41421356237310$$ $$b_2\approx 1.53208888623796$$ The first one is clearly $b_1=\sqrt 2$; for the second root, an inverse symbolic calculator reports that this is a solution of the cubic $x^3-3x+1=0$. Using Cardano method, this equation presents three real roots. So, using the trigonometric method for solving cubic equations, the only acceptable root is given by $$b_2=\sqrt{3} \sin \left(\frac{\pi }{9}\right)+\cos \left(\frac{\pi }{9}\right)=2 \cos \left(\frac{2 \pi }{9}\right)$$ Back to the definition of $\cos(x)$ as a function of $b$, we then find $$b_1=\sqrt 2\implies x_1=\frac{\pi }{3}$$ $$b_2=2 \cos \left(\frac{2 \pi }{9}\right)\implies x_2=\frac{\pi }{6}$$ However, $x_1$ satifies the first equation but not the second (this probably corresponds to a false root introduced by the multiple quaring processes). 
So $x_2=\frac{\pi }{6}$ is the only solution.
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