<s>[INST] Solve for $x: 2/\sqrt{2-x^2} = 4- 2x^2?$ How do you expand and solve for $x$?
It is $2 = 4\sqrt{2-x^2} - 2x^2\sqrt{2-x^2}$
Thank you!
How would I solve for x?
 [/INST] Note that real solutions of $x$ occur for $|x|\lt\sqrt 2$.  Then start by dividing by $2$:
$$1=(2-x^2)\sqrt{2-x^2}=(2-x^2)^{3/2}$$
$$\implies 1=2-x^2 \implies x^2=1$$
Which implies that $x=\pm 1$.
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