Question Number 196885 by hardmath last updated on 02/Sep/23
Answered by MM42 last updated on 02/Sep/23
$$\mathrm{2}{cosx}=\sqrt{\mathrm{2}+\mathrm{2}{cos}\mathrm{2}{x}}=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+{cos}\mathrm{4}{x}}} \\ $$$$=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+{cos}\mathrm{8}{x}}}}=….=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+…}}} \\ $$$${let}\:\:{x}=\mathrm{0}\Rightarrow\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+…}}}=\mathrm{2}\:\checkmark \\ $$$$ \\ $$
Commented by hardmath last updated on 02/Sep/23
$$\mathrm{cool}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{professor} \\ $$
Commented by hardmath last updated on 03/Sep/23
$$\mathrm{Sorry}\:\mathrm{professor},\:\mathrm{but}\:\mathrm{equalities}\:\mathrm{are}\:\mathrm{satisfied} \\ $$$$\mathrm{only}\:\mathrm{if}\:\:\mathrm{cos}\left(\left(\mathrm{2}^{\mathrm{n}} \right)\mathrm{x}\right)\geqslant\mathrm{0}\:\:\forall\mathrm{n}\geqslant\mathrm{0}\:\:\mathrm{so}\:\:\left(-\:\frac{\pi}{\mathrm{2}}\right)^{\mathrm{n}+\mathrm{1}} \leqslant\:\mathrm{x}\:\leqslant\:\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{n}+\mathrm{1}} \:\:\forall\mathrm{n}\geqslant\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{x}\:=\:\mathrm{0} \\ $$$$\mathrm{Therefore}\:\:\mathrm{2cos}\left(\mathrm{0}\right)\:=\:?\:\Rightarrow\:\mathrm{2}\:=\:? \\ $$
Answered by Mathspace last updated on 02/Sep/23
$${x}=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}+\sqrt{…}}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\mathrm{2}+{x}\:\Rightarrow{x}^{\mathrm{2}} −{x}−\mathrm{2}=\mathrm{0} \\ $$$$\Delta=\mathrm{1}+\mathrm{8}=\mathrm{9}\:\Rightarrow \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{1}+\sqrt{\mathrm{9}}}{\mathrm{2}}\:\:=\mathrm{2} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{1}−\sqrt{\mathrm{9}}}{\mathrm{2}}=−\mathrm{1} \\ $$$${but}\:{x}>\mathrm{0}\:\Rightarrow{x}=\mathrm{2} \\ $$