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Question Number 17150 by Tinkutara last updated on 01/Jul/17
If the equation cos x + 3 cos (2Kx) = 4 has exactly one solution, then (1) K is a rational number of the form (P/(P + 1)), P ≠ −1 (2) K is irrational number whose rational approximation does not exceed 2 (3) K is irrational number (4) K is a rational number of the form (P/(P − 1)), P ≠ 1
$$\mathrm{If}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{cos}\:{x}\:+\:\mathrm{3}\:\mathrm{cos}\:\left(\mathrm{2}{Kx}\right)\:=\:\mathrm{4} \\ $$$$\mathrm{has}\:\mathrm{exactly}\:\mathrm{one}\:\mathrm{solution},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:{K}\:\mathrm{is}\:\mathrm{a}\:\mathrm{rational}\:\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form} \\ $$$$\frac{{P}}{{P}\:+\:\mathrm{1}},\:{P}\:\neq\:−\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:{K}\:\mathrm{is}\:\mathrm{irrational}\:\mathrm{number}\:\mathrm{whose} \\ $$$$\mathrm{rational}\:\mathrm{approximation}\:\mathrm{does}\:\mathrm{not} \\ $$$$\mathrm{exceed}\:\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:{K}\:\mathrm{is}\:\mathrm{irrational}\:\mathrm{number} \\ $$$$\left(\mathrm{4}\right)\:{K}\:\mathrm{is}\:\mathrm{a}\:\mathrm{rational}\:\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form} \\ $$$$\frac{{P}}{{P}\:−\:\mathrm{1}},\:{P}\:\neq\:\mathrm{1} \\ $$
Commented by prakash jain last updated on 03/Jul/17
cos x+3cos (2Kx)=4 cos x≤1 cos x=1 cos 2Kx=1 x=2nπ 2mx=4Knπ 4Knπ=2mπ n has only one solution. if K is rational (p/q) n=((2mq)/(4p))=((mq)/p) has infinite solution for m=lp so K is not rational. K irrational n=(m/(2K)) gives only one solution m=0,n=0
$$\mathrm{cos}\:{x}+\mathrm{3cos}\:\left(\mathrm{2}{Kx}\right)=\mathrm{4} \\ $$$$\mathrm{cos}\:{x}\leqslant\mathrm{1} \\ $$$$\mathrm{cos}\:{x}=\mathrm{1} \\ $$$$\mathrm{cos}\:\mathrm{2}{Kx}=\mathrm{1} \\ $$$${x}=\mathrm{2}{n}\pi \\ $$$$\mathrm{2}{mx}=\mathrm{4}{Kn}\pi \\ $$$$\mathrm{4}{Kn}\pi=\mathrm{2}{m}\pi\: \\ $$$${n}\:{has}\:{only}\:{one}\:{solution}. \\ $$$${if}\:{K}\:{is}\:{rational}\:\frac{{p}}{{q}} \\ $$$${n}=\frac{\mathrm{2}{mq}}{\mathrm{4}{p}}=\frac{{mq}}{{p}}\:{has}\:{infinite}\:{solution} \\ $$$${for}\:{m}={lp} \\ $$$$\mathrm{so}\:{K}\:\mathrm{is}\:\mathrm{not}\:\mathrm{rational}. \\ $$$${K}\:\mathrm{irrational} \\ $$$${n}=\frac{{m}}{\mathrm{2}{K}}\:\mathrm{gives}\:\mathrm{only}\:\mathrm{one}\:\mathrm{solution} \\ $$$${m}=\mathrm{0},{n}=\mathrm{0} \\ $$
Commented by 1234Hello last updated on 03/Jul/17
But why 2Kx = 4Knπ?
$$\mathrm{But}\:\mathrm{why}\:\mathrm{2}{Kx}\:=\:\mathrm{4}{Kn}\pi? \\ $$
Commented by prakash jain last updated on 03/Jul/17
Typo It should be 2mx. x=2nπ for cos x=1 for cos 2Kx=1⇒2Kx=2mπ x=2nx 4Knx=2mπ
$$\mathrm{Typo} \\ $$$$\mathrm{It}\:\mathrm{should}\:\mathrm{be}\:\mathrm{2}{mx}. \\ $$$${x}=\mathrm{2}{n}\pi\:\mathrm{for}\:\mathrm{cos}\:{x}=\mathrm{1} \\ $$$${for}\:\mathrm{cos}\:\mathrm{2}{Kx}=\mathrm{1}\Rightarrow\mathrm{2}{Kx}=\mathrm{2}{m}\pi \\ $$$${x}=\mathrm{2}{nx} \\ $$$$\mathrm{4}{Knx}=\mathrm{2}{m}\pi \\ $$

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