idiopatia0f

2022-01-14

Equation $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$
Using $x=2\mathrm{cos}\left(\theta \right)$ we get $2\mathrm{cos}\left(\theta \right)=\sqrt{2+\sqrt{2-\sqrt{2+2\mathrm{cos}\theta }}}$
$2\mathrm{cos}\theta =\sqrt{2+\sqrt{2-2\mathrm{cos}\frac{\theta }{2}}}$
$2\mathrm{cos}\theta =\sqrt{2+2\mathrm{sin}\frac{\theta }{4}}$
$4{\mathrm{cos}}^{2}\theta =2+2\mathrm{sin}\frac{\theta }{4}$
After this step I am not able to solve it

David Clayton

My illustration is outlined below
$4{\mathrm{cos}}^{2}\theta =2\left(1+\mathrm{sin}\frac{\theta }{4}\right)$
$4{\mathrm{cos}}^{2}{10}^{\circ }\ne 2\left(1+\mathrm{sin}\frac{{10}^{\circ }}{4}\right)$
$4{\mathrm{cos}}^{2}{10}^{\circ }\ne 2\left(1+{\mathrm{sin}2.5}^{\circ }\right)$
$4{\mathrm{cos}}^{2}{10}^{\circ }\ne 2\left(1+{\mathrm{cos}87.5}^{\circ }\right)$
$4{\mathrm{cos}}^{2}{20}^{\circ }\ne 2\left(1+\mathrm{sin}\frac{{10}^{\circ }}{4}\right)$
$4{\mathrm{cos}}^{2}{20}^{\circ }\ne 2\left(1+{\mathrm{sin}5}^{\circ }\right)$
$4{\mathrm{cos}}^{2}{20}^{\circ }\ne 2\left(1+{\mathrm{cos}85}^{\circ }\right)$
$4{\mathrm{cos}}^{2}{40}^{\circ }\ne 2\left(1+\mathrm{sin}\frac{{40}^{\circ }}{4}\right)$
$4{\mathrm{cos}}^{2}{40}^{\circ }\ne 2\left(1+{\mathrm{sin}10}^{\circ }\right)$
$4{\mathrm{cos}}^{2}{40}^{\circ }\ne 2\left(1+{\mathrm{cos}80}^{\circ }\right)$
$4{\mathrm{cos}}^{2}{80}^{\circ }\ne 2\left(1+\mathrm{sin}\frac{{80}^{\circ }}{4}\right)$
$4{\mathrm{cos}}^{2}{80}^{\circ }\ne 2\left(1+{\mathrm{sin}20}^{\circ }\right)$
$4{\mathrm{cos}}^{2}{80}^{\circ }\ne 2\left(1+{\mathrm{cos}70}^{\circ }\right)$
Hence the option is $0={40}^{\circ }$

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