Very difficult equation

by MM » Thu Jan 29, 2009 3:40 pm

Solve in real numbers the equation
$\-x+sqrt{1-x^{2}}\-=sqrt{2}\left(2x^{2}-1\right)$ .

by **srun** » Fri Feb 06, 2009 4:59 pm

See attachment.[/img]

by **martosss** » Sat May 09, 2009 10:18 am

$Substitute\; x=cos\alp, \alp\in[0: ;\: \pi ]$
$Then\; |x+\sqrt{1-x^2}|=\sqrt{2}(2x^2-1)\Leftright |cos\alp +sin\alp |=\sqrt{2}(2cos^2\alp -1)$
$|\N {\sqrt{2}}cos(\alp-\frac{\pi}{4})|=\N {\sqrt{2}}cos(2\alp )\Right \alp\in[0\: ;\: \frac{\pi}{4}]\cup [\frac{3\pi}{4}\: ;\: \pi]$
1) $\alp \in [0\: ;\: \frac{\pi}{4}]$
$cos(\alp -\frac{\pi}{4})=cos(2\alp )\dots$
2. $\alp\in [\frac{3\pi}{4}\: ;\: \pi]$
$-cos(\alp -\frac{\pi}{4})=cos(2\alp )\dots$

Very difficult equation

Very difficult equation

Solution

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