<s>[INST] Solving a Radical Equation $5(\sqrt{1-x} + \sqrt{1+x}) = 6x + 8\sqrt{1-x^2}$ (squaring doesn't help) How should I approach this problem:
$$ 5(\sqrt{1-x} + \sqrt{1+x}) = 6x + 8\sqrt{1-x^2} $$
I've tried squaring both sides but to get rid of all the radicals requires turning it into a quartic equation, which I don't know how to solve?
Any tips? Thanks!
 [/INST] HINT:
Method$\#1:$
Let $2y=\arccos x\implies x=\cos2y,\sqrt{1-x^2}=+|\sin2y|$
As for real $a,\sqrt a\ge0,$
Using the definition of Principal values, $0\le2y\le\pi\implies\sqrt{1-x^2}=+\sin2y$
Again, $0\le2y\le\pi\iff0\le y\le\dfrac\pi2\implies\sin y,\cos y\ge0$
$\implies\sqrt{1-x}=+\sqrt2\sin y,\sqrt{1+x}=+\sqrt2\cos y$
Method$\#2:$
Let $x=\cos2y$
$\sqrt{1-x^2}=|\sin2y|$
$1-x=2\sin^2y\implies\sqrt{1-x}=\sqrt2|\sin y|$ and simialrly $\sqrt{1+x}=\sqrt2|\cos y|$
Now for real $a, |a|=+a$ if $a\ge0,$ else $-a$
Case $\#1:$
$\sin y,\cos y\ge0\implies\sin2y=2\sin y\cos y\ge0$
Case $\#2:$
$\sin y<0,\cos y<0\implies\sin2y=\cdots>0$
Case $\#3:$
$\sin y>0,\cos y<0\implies\sin2y=\cdots$
Case $\#4:$
$\sin y<0,\cos y>0\implies\sin2y=\cdots$
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