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!
 A: 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 $\#4:$
$\sin y<0,\cos y>0\implies\sin2y=\cdots$
A: after squaring two times and sorting and factorizing i got this here
$$-4 (25 x-24) \left(100 x^3-75 x+24\right)=0$$
A: Here is a really nice method: when looking at terms such as $\sqrt{1-x}$, $\sqrt{1+x}$, and $\sqrt{1-x^2}$, think trigonometry!
Make the trig substitution:
$$x = \cos{\theta}$$
Now:
$$\sqrt{1-x}=\sqrt{2}\sin{\frac{\theta}{2}}$$
$$\sqrt{1+x}=\sqrt{2}\cos{\frac{\theta}{2}}$$
$$\sqrt{1-x^2}=\sin{\theta}$$
The equation with $\theta$ is then:
$$5\left(\sin{\frac{\theta}{2}} + \cos{\frac{\theta}{2}}\right) = 6\cos{\theta} + 8\sin{\theta} $$
Can you take it from here?
Edit: Another trick is used to finish the problem. One of the advantages of trig substitutions is that there are many identities to use. In this case:
$$6\cos{\theta} + 8\sin{\theta} = 10\left(\frac{3}{5}\cos{\theta} + \frac{4}{5}\sin{\theta}\right)$$
Now, if we let $\phi = \sin^{-1}\left(\frac{3}{5}\right)$:
$$6\cos{\theta} + 8\sin{\theta} = 10\left(\sin{\phi}\cos{\theta} + \cos{\phi}\sin{\theta}\right)=10\sin(\phi+\theta)$$
