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

Question

I have to solve this irrational equation on $\mathbb{R}$ : $$ \sqrt{1-x}=2x^2-1+2x\sqrt{1-x^2}$$

I tried to do a substitution with $u=1-x$ but the only things I manage to reach is the following equation by squaring and using $(a-b)(a+b)=a^2 -b^2$: $$ (\sqrt{1-x}-2x\sqrt{1-x^2})^2 = (2x^2 -1)^2$$ $$\implies 1 - x + 4 x^2 - 4 x^4 - 4x \sqrt{1 - x} \sqrt{1 - x^2} = 4x^4 - 4x^2 +1$$ $$ \implies -4x\sqrt{(1-x)(1+x)(1-x)} = 8x^4-8x^2$$ $$ \implies 4(1-x)\sqrt{1+x} = 8x^4 -8x^2$$ $$ \implies (1-x)\sqrt{1+x} = 2x^2 (1-x^2)$$

I don't manage to go forward. The only thing I know is that the solution (if there is one) is in [-1;1].

Could you help me, please ?

Almost universal rule: if the substitution $u=1-x$ simplifies the problem, then so did the substitution $u=x$. — , Sep 05 '20 at 14:34

Michael Rozenberg · Accepted Answer · 2020-09-06T08:15:59.597

14

Let $x=\cos\alpha,$ where $\alpha\in[0,\pi]$.

Thus, we need to solve $$\sqrt{1-\cos\alpha}=2\cos^2\alpha-1+2\cos\alpha\sqrt{1-\cos^2\alpha}$$ or $$\sqrt{2\sin^2\frac{\alpha}{2}}=\cos2\alpha+2\cos\alpha\sin\alpha$$ $$\sqrt2\sin\frac{\alpha}{2}=\cos2\alpha+\sin2\alpha$$ or $$\sin\frac{\alpha}{2}=\sin\left(\frac{\pi}{4}+2\alpha\right),$$ which gives $$\frac{\alpha}{2}=\frac{\pi}{4}+2\alpha+2\pi k,$$ where $k\in\mathbb Z,$ which is impossible for $\alpha\in[0,\pi]$ or $$\frac{\alpha}{2}=\pi-\left(\frac{\pi}{4}+2\alpha\right)+2\pi k,$$ where $k\in\mathbb Z,$ which is possible for $k=0$ only, which gives $$\alpha=\frac{3\pi}{10}$$ and the root $$\cos54^{\circ}$$ or $$\sin36^{\circ}$$ or $$\frac{\sqrt{5-\sqrt5}}{2\sqrt2}.$$

edited Sep 06 '20 at 08:15

answered Sep 05 '20 at 14:27

Michael Rozenberg

203,855

OOOH ! Thanks a lot ! The interval [-1;1] would had been help me. I didn't manage to think out of the box. So the answer is sin(\pi / 6 ) =1/2. But I don't figure out how you reached those two equations – Hugo Faurand Sep 05 '20 at 14:51
2

@Hugo Faurand Add please another attempts in your starting post. For example, how you got your equation, which you wrote after your trying? Show, how you made squaring. Do it please. Otherwise, your topic would be deleted and I don't want that my solution would be deleted. – Michael Rozenberg Sep 05 '20 at 14:54
@Hugo Faurand I got the following answer: $\sin36^{\circ}$. – Michael Rozenberg Sep 05 '20 at 15:52
Can you explain briefly how you got the equation with the √2 factor ? As far as I'm concerned I solved the system (a/2 =π/4+2a. ; a/2=π-π/4-2a) – Hugo Faurand Sep 05 '20 at 19:41
@Hugo Faurand I used $1-\cos\alpha=2\sin^2\frac{\alpha}{2}.$ – Michael Rozenberg Sep 05 '20 at 19:43
I don't manage to reach your equations. Do you do the substitution in the last step of my first post or directly in the initial equation ? – Hugo Faurand Sep 06 '20 at 07:47
@Hugo Faurand I added something. See now. Ask your questions. – Michael Rozenberg Sep 06 '20 at 08:16
Let us continue this discussion in chat. – Hugo Faurand Sep 06 '20 at 13:07

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

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