Solve the equation $\sqrt{1-x}=2x^2-1+2x\sqrt{1-x^2}$ Solve the following equation: $\sqrt{1-x}=2x^2-1+2x\sqrt{1-x^2}$
Unfortunately I have no idea.
 A: Put $x=\cos \theta$, then:
$\sqrt{1-\cos \theta} = 2\cos^2 \theta -1 \pm 2\cos \theta\sin\theta = \cos 2\theta \pm \sin 2\theta$.
Squaring ,we get:
$1-\cos \theta = 1 \pm \sin 4\theta$.
Therefore $\cos \theta = \pm\sin 4\theta$. The solutions of this equation are easy to find. Also, one must eliminate extraneous solutions e.g. $126^\circ$.
A: First notice that $x$ must be in $[-1,1]$. Then rewrite $$\sqrt{1-x}-2x\sqrt{1-x^2} =2x^2-1$$ and square : $$(1-x)+4x^2(1-x^2)-4x(1-x)\sqrt{1+x}=4x^4-4x^2+1.$$ This can be simplified $$4x(x-1)\sqrt{1+x}=8x^4-8x^2+x=x(8x^3-8x+1).$$ Symplify by $x$ and square another time to get $$16(x+1)(x-1)^2=64x^6+64x^2+1+16x^3-16x-128x^4.$$ Finally this leads to the equation $$64x^6-128x^4+80x^2-15 = 0.$$ Set $y=x^2$ and solve $$64y^3-128y^2+80y-15=0.$$ An "obvious solution" is $\frac{3}{4}$. After that you will have to solve a simple second-degree equation.
Don't forget to discuss the solutions at the end. If no mistakes, only one solution should be in $[-1,1]$ and satisfy the original equation.
A: Let $x=\cos2y$
WLOG $0\le2y\le180^\circ\implies\sin y\ge0$
$$\sqrt2\sin y=\cos4y+\sin4y\iff\sin\left(4y+45^\circ\right)=\sin y$$
$\implies$
either $4y+45^\circ=360^\circ n+y\iff y=120^\circ n-15^\circ\implies y=(120-15)^\circ\implies x=\cos2y=?$
or $4y+45^\circ=(2n+1)180^\circ-y\iff y=72^\circ n+27^\circ\implies y=27^\circ\implies x=\cos2y=?$
