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Question Number 17150 by Tinkutara last updated on 01/Jul/17

$$\mathrm{If}\mathrm{the}\mathrm{equation}\cos x+3\cos \left(2Kx\right)=4$$$$\mathrm{has}\mathrm{exactly}\mathrm{one}\mathrm{solution},\mathrm{then}$$$$\left(1\right)K\mathrm{is}\mathrm{a}\mathrm{rational}\mathrm{number}\mathrm{of}\mathrm{the}\mathrm{form}$$$$\frac{P}{P+1},P\ne -1$$$$\left(2\right)K\mathrm{is}\mathrm{irrational}\mathrm{number}\mathrm{whose}$$$$\mathrm{rational}\mathrm{approximation}\mathrm{does}\mathrm{not}$$$$\mathrm{exceed}2$$$$\left(3\right)K\mathrm{is}\mathrm{irrational}\mathrm{number}$$$$\left(4\right)K\mathrm{is}\mathrm{a}\mathrm{rational}\mathrm{number}\mathrm{of}\mathrm{the}\mathrm{form}$$$$\frac{P}{P-1},P\ne 1$$

Commented by prakash jain last updated on 03/Jul/17

$$\cos x+3\cos \left(2Kx\right)=4$$$$\cos x⩽1$$$$\cos x=1$$$$\cos 2Kx=1$$$$x=2n\pi$$$$2mx=4Kn\pi$$$$4Kn\pi =2m\pi$$$$nhasonlyonesolution.$$$$ifKisrational\frac{p}{q}$$$$n=\frac{2mq}{4p}=\frac{mq}{p}hasinfinitesolution$$$$form=lp$$$$\mathrm{so}K\mathrm{is}\mathrm{not}\mathrm{rational}.$$$$K\mathrm{irrational}$$$$n=\frac{m}{2K}\mathrm{gives}\mathrm{only}\mathrm{one}\mathrm{solution}$$$$m=0,n=0$$

Commented by 1234Hello last updated on 03/Jul/17

$$\mathrm{But}\mathrm{why}2Kx=4Kn\pi ?$$

Commented by prakash jain last updated on 03/Jul/17

$$\mathrm{Typo}$$$$\mathrm{It}\mathrm{should}\mathrm{be}2mx.$$$$x=2n\pi \mathrm{for}\cos x=1$$$$for\cos 2Kx=1\Rightarrow 2Kx=2m\pi$$$$x=2nx$$$$4Knx=2m\pi$$

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