Why this system have one solution

Question

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.

Answer 1

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 $x=π/6$

Answer 2

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.

Answer 3

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.