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Question Number 196885 by hardmath last updated on 02/Sep/23

Answered by MM42 last updated on 02/Sep/23

$$2cosx=\sqrt{2+2cos2x}=\sqrt{2+\sqrt{2+cos4x}}$$$$=\sqrt{2+\sqrt{2+\sqrt{2+cos8x}}}=....=\sqrt{2+\sqrt{2+\sqrt{2+...}}}$$$$letx=0\Rightarrow \sqrt{2+\sqrt{2+\sqrt{2+...}}}=2✓$$$$$$

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

$$\mathrm{Sorry}\mathrm{professor},\mathrm{but}\mathrm{equalities}\mathrm{are}\mathrm{satisfied}$$$$\mathrm{only}\mathrm{if}\cos \left(\left(2^{\mathrm{n}}\right)\mathrm{x}\right)⩾0\mathrm{\forall }\mathrm{n}⩾0\mathrm{so}{(-\frac{\pi }{2})}^{\mathrm{n}+1}⩽\mathrm{x}⩽{\left(\frac{\pi }{2}\right)}^{\mathrm{n}+1}\mathrm{\forall }\mathrm{n}⩾0$$$$\Rightarrow \mathrm{x}=0$$$$\mathrm{Therefore}2\cos \left(0\right)=?\Rightarrow 2=?$$

Answered by Mathspace last updated on 02/Sep/23

$$x=\sqrt{2+\sqrt{2+\sqrt{2}+\sqrt{...}}}$$$$\Rightarrow x^2=2+x\Rightarrow x^2-x-2=0$$$$\mathrm{\Delta }=1+8=9\Rightarrow$$$$x_1=\frac{1+\sqrt{9}}{2}=2$$$$x_2=\frac{1-\sqrt{9}}{2}=-1$$$$butx>0\Rightarrow x=2$$

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