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I need to solve the equation $$9x\sqrt{1-4x^2}+2x\sqrt{1-81x^2}=\sqrt{1-144x^2}$$ over real numbers.

Squaring both sides and transforming this in 4-degree equation I got $x = \frac{1}{16}$. But can we solve this equation in an easier way?

The fact that $|x|<1$, as well as presence of square roots of the difference of squares, suggests that perhaps trigonometry needs to be applied here, but I don’t know how.

Ivan Borisyuk
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    Are you sure that $x=\frac1{16}$? WolframAlpha shows different results, ie that $\frac1{16}$ is just an approximate answer. – Sahaj Jun 16 '24 at 18:07
  • Elimination of all square roots yields $$1 - 458 x^2 + 53251 x^4 - 365634 x^6 + 164025 x^8 == 0$$ with terrible expression for the root of $x^2$ – Roland F Jun 16 '24 at 18:32
  • Sorry, I wrote the equation incorrectly, the correct equation is $9x\sqrt{1-4x^2} + 2x\sqrt{1-81x^2} = \sqrt{1-144x^2}$ – Ivan Borisyuk Jun 16 '24 at 19:23

2 Answers2

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We can indeed exploit some trigonometric identities (terms and conditions apply):

$$\begin{align*} \arcsin\left(x\sqrt{1-y^2} + y \sqrt{1-x^2}\right) &= \arcsin x + \arcsin y \tag{$*$} \\ \arcsin\sqrt{1-x^2} = \arccos x &= \frac\pi2 - \arcsin x \end{align*}$$

Together, these let us rewrite the equation as

$$\arcsin a + \arcsin b + \arcsin c = \frac\pi2$$

where $(a,b,c)=(2x,9x,12x)$. Further application of $(*)$ allows us to recover more equations,

$$\begin{align*} a\sqrt{1-b^2} + b\sqrt{1-a^2} &= \sqrt{1-c^2} & (\rm OP)\\ b\sqrt{1-c^2} + c\sqrt{1-b^2} &= \sqrt{1-a^2} \\ a\sqrt{1-c^2} + c\sqrt{1-a^2} &= \sqrt{1-b^2} \end{align*}$$

By substitution, e.g. $\sqrt{1-b^2}$ from the third equation into the first, we can recover (one of) the suggested solution(s),

$$\begin{align*} a\left(a\sqrt{1-c^2} + c\sqrt{1-a^2}\right) + b\sqrt{1-a^2} &= \sqrt{1-c^2} \\ (ac+b) \sqrt{1-a^2} &= \sqrt{1-c^2} (1-a^2) \\ ac+b &= \sqrt{1-a^2} \sqrt{1-c^2} \\ (ac+b)^2 &= (1-a^2)(1-c^2) \\ a^2 + b^2 + c^2 &= 1 - 2abc \end{align*}$$

$$\implies 432x^3 + 229x^2 - 1 = 0$$

$x=\dfrac1{16}$ is a solution to this cubic. One can use the rational root theorem to deduce this, though I suspect there's a more efficient method.

user170231
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  • +1: One can also show that $x>0$ (otherwise the equation in question is +ve on one side and -ve on the other), and that there is exactly at most one solution (which is by comparing that the LHS is increasing after 0 and the RHS is decreasing). – Sahaj Jun 17 '24 at 20:03
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    Is there a typo in the last line? – John Bentin Jun 17 '24 at 20:22
  • @JohnBentin Fixed. Thanks – user170231 Jun 17 '24 at 20:34
  • For completeness, it should be noted that the other two roots of the cubic are $x = \frac{-8 \pm \sqrt{37}}{27}$. – Dan Jun 17 '24 at 21:54
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Following your suggestion, in the given equation $$9x\sqrt{1-4x^2}+2x\sqrt{1-81x^2}=\sqrt{1-144x^2},$$ where we note that $0<x<\frac1{12}$, let us write $$\cos\theta=2x\;\;\text{and}\;\; \cos\phi=9x,\quad\text{with}\quad\theta,\phi\in(0\,\pmb,\,\tfrac12\pi).\qquad\quad(1)$$

The original equation becomes $\sin(\theta+\phi)=\sqrt{1-8\cos\theta\cos\phi}$. Squaring then gives $$\cos^2(\theta+\phi)=8\cos\theta\cos\phi=36\cos^2\theta.$$ Now $x<\frac1{12},\cos\theta<\frac16,\cos\phi<\frac34,$ and so $\cos(\theta+\phi)<0$. Hence $\cos(\theta+\phi)=-6\cos\theta,$ or $$\cos\theta\cos\phi+6\cos\theta=\sin\theta\sin\phi.$$ Squaring now yields $(6+\cos\phi)^2\cos^2\theta=(1-\cos^2\theta)(1-\cos^2\phi)$. In terms of the original variable (see eqn $1$), this reduces to a cubic equation in $x$: $$432x^3+229x^2-1=0.\qquad\quad(2)$$ All the numbers hitherto have been powers of $2$ or $3$, or products of them, but now the ugly number $229$ pops up. Can we do something about that? Well, notice that $229=16^2-27$, while $432=16\times27$. Conveniently, that allows the factorization of eqn $2$ as $$(16x-1)(27x^2+16x+1)=0.$$ The quadratic factor is always positive, so the only solution is $x=\frac1{16}$.

John Bentin
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