Tough Irrational Equation highschool

Question

Have been trying to solve this irrational equation for a day but as it seems, i'm not going anywere with it. Can somebody offer me a tip ? Thanks!

*Tried a "t" substitution for x squared but it still yields a 4th degree polynomial equation instead of an 8th, which I think can be solved by factoring, but i'm sure there's an easier way to do it.

$$\sqrt{\frac{1+2x\sqrt{1-x^2}}{2}} + 2x^2 =1$$

Answer 1

We have $$\sqrt{\frac{x^2+2x\sqrt{1-x^2}+1-x^2}{2}}+2x^2-1=0$$ or $$\frac{|x+\sqrt{1-x^2}|}{\sqrt2}+2x^2-1=0.$$ Now, $$x=\sqrt{1-x^2}$$ gives $x=\frac{1}{\sqrt2}$ which is not a root of our equation.

Thus, our equation is equivalent to $$\frac{|2x^2-1|}{\sqrt2|x-\sqrt{1-x^2}|}+2x^2-1=0$$ or since $2x^2-1\leq0,$ $$(2x^2-1)\left(1-\frac{1}{\sqrt2|x-\sqrt{1-x^2}|}\right)=0,$$ which gives $$x=-\frac{1}{\sqrt2}$$ or $$\sqrt2|x-\sqrt{1-x^2}|=1.$$ Now, if $x-\sqrt{1-x^2}\geq0$ we obtain $x\geq0$ and $x^2\geq1-x^2\geq0$ or $2x^2-1\geq0,$

which with $2x^2-1\leq0$ gives $x=\pm\frac{1}{\sqrt2},$ which is impossible.

Id est, $x-\sqrt{1-x^2}<0$ and it remains to solve $$\sqrt2(x-\sqrt{1-x^2})=-1.$$ Can you end it now?

Answer 2

Hint Set $x =\sin(t)$ for some $t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. Then $$\sqrt{\frac{1+2x\sqrt{1-x^2}}{2}} + 2x^2 =1\\ \sqrt{\frac{1+2\sin(t)\cos(t)}{2}} =1-2 \sin^2(t) \\ \sqrt{\frac{1+\sin(2t)}{2}} =\cos(2t) \\ \sqrt{\frac{1+\cos(\frac{\pi}{2}-2t)}{2}} =\cos(2t) \\ \pm \cos(\frac{\pi}{4}-t) =\cos(2t) \\ $$