<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] Solving this is mainly just fiddly algebra.
$$2b^2+b-4=2\sqrt{4-b^2}\cos{x}$$
$$2b^2-4=2b\cos{(x+\frac{\pi}{18})}-2\sqrt{4-b^2}\cos{\frac{5\pi}{18}}$$
The second can be rewritten as:
$$2b^2-4=2b\left(\cos x\cos\frac{\pi}{18}-\sin x\sin\frac{\pi}{18}\right)-2\sqrt{4-b^2}\cos{\frac{5\pi}{18}}$$
Rearranging the first one gives:
$$\cos x=\frac{2b^2+b-4}{2\sqrt{4-b^2}}$$
and hence:
$$\sin x=\sqrt{1-\frac{(2b^2+b-4)^2}{4(4-b^2)}}=\frac{\sqrt{4(4-b^2)-(2b^2+b-4)^2}}{2\sqrt{4-b^2}}$$
Putting this together gives:
$$2b^2-4=2b\left(\frac{2b^2+b-4}{2\sqrt{4-b^2}}\cos\frac{\pi}{18}-\frac{\sqrt{4(4-b^2)-(2b^2+b-4)^2}}{2\sqrt{4-b^2}}\sin\frac{\pi}{18}\right)-2\sqrt{4-b^2}\cos{\frac{5\pi}{18}}$$
By repeated rearranging and squaring you can get rid of all the square roots and eventually end up with a high degree polynomial in $b$ to solve. It is unlikely to factorize nicely. Numerically technique would most likely be required to determine the roots of the equation.
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