Solve the following equation: $\sqrt{1-x}=2x^2-1+2x\sqrt{1-x^2}$
Unfortunately I have no idea.
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Wolfram gives a single solution. If anyone has an account they can check their step by step method. Given the nested radicals it might be messy. http://www.wolframalpha.com/input/?i=%5Csqrt%7B1-x%7D%3D2x%5E2-1%2B2x%5Csqrt%7B1-x%5E2%7D – pancini Apr 14 '16 at 07:41
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1One idea which comes to mind is substituting $x = \sin t$. Not sure if that leads anywhere, though. – MathematicsStudent1122 Apr 14 '16 at 07:41
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This is a fierce algebraic problem... If you try to eliminate radicals we end up with an 8th degree polynomial. – Sidharth Ghoshal Apr 14 '16 at 07:44
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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. $126^\circ$.
Sarvesh Ravichandran Iyer
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@ElliotG Thank you, but credit must go to Mathematicsstudent1122 above. – Sarvesh Ravichandran Iyer Apr 14 '16 at 08:03
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We have $\sin (-x) = -\sin x$, so shouldn't we have $\sin(90-\theta)=-\sin(\theta-90)$ instead of $\sin(90-\theta)=-\sin(90+\theta)$? – Ovi Apr 14 '16 at 16:34
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@Ovi Exactly, and then $\sin(\theta-90)$ is $\sin(\theta-90+180)$, because that is period of the sine function, and that gives $\sin(90+\theta)$, which is what I wrote on top. Please clarify if you continue to have doubts. – Sarvesh Ravichandran Iyer Apr 15 '16 at 03:47
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I thought the period of sine was $360$, not $180$. $\sin (x+180)$ seems to equal $-\sin x$ since $\sin(x+180)=\sin x \cos 180+ \sin 180 \cos x = -\sin x$, so $\sin(90-x)=-\sin(x-90)=-(-\sin(x-90+180))=\sin(90+x)$, whereas you got a $-\sin(90+x)$ – Ovi Apr 15 '16 at 15:37
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You're working in degree measure, so you should have $180n$ not $\pi n$. This also means that 126 degrees is not a solution. – Semiclassical Apr 23 '16 at 01:12
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I will edit my answer, @Semiclassical. Thank you for pointing this out. – Sarvesh Ravichandran Iyer Apr 23 '16 at 01:13
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@Ovi Thank you for pointing out my mistake. I have corrected it, and provided a little more to the answer. Now it should be amply clear where my answer comes from. – Sarvesh Ravichandran Iyer Apr 23 '16 at 01:20
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@астонвіллаолофмэллбэрг, If $x=\cos126^\circ,$ $$2x^2-1, x<0$$ The Right Hand Side $$=-\sqrt2\sin(45+72)^\circ=-\sqrt2\sin63^\circ$$ and the left $$=+\sqrt2\sin63^\circ$$ – lab bhattacharjee Apr 23 '16 at 11:53
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@RFZ, http://www.mathwords.com/p/periodicity_identities.htm and http://mathsfirst.massey.ac.nz/Trig/TrigGenSol.htm – lab bhattacharjee Apr 23 '16 at 11:58
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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.
C. Dubussy
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Of course, I didn't write it, but it's ok since $0$ is clearly not a solution of the initial problem. – C. Dubussy Apr 14 '16 at 08:11
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That has three solutions, $y=\frac{3}{4},\frac{5\pm\sqrt5}{8}$. But only $x=\sqrt{\frac{5-\sqrt5}{8}}$ satisfies the original equation. – almagest Apr 14 '16 at 08:12
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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 $4y+45^\circ=(2n+1)180^\circ-y\iff y=72^\circ n+27^\circ\implies y=27^\circ\implies x=\cos2y=?$
lab bhattacharjee
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