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

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
$$\frac{\sqrt{5-\sqrt5}}{2\sqrt2}.$$
