<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] 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 $\phi = \sin^{-1}\left(\frac{3}{5}\right)$:
$$6\cos{\theta} + 8\sin{\theta} = 10\left(\sin{\phi}\cos{\theta} + \cos{\phi}\sin{\theta}\right)=10\sin(\phi+\theta)$$
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