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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
Iftheequationcosx+3cos(2Kx)=4hasexactlyonesolution,then(1)KisarationalnumberoftheformPP+1,P1(2)Kisirrationalnumberwhoserationalapproximationdoesnotexceed2(3)Kisirrationalnumber(4)KisarationalnumberoftheformPP1,P1
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
cosx+3cos(2Kx)=4cosx1cosx=1cos2Kx=1x=2nπ2mx=4Knπ4Knπ=2mπnhasonlyonesolution.ifKisrationalpqn=2mq4p=mqphasinfinitesolutionform=lpsoKisnotrational.Kirrationaln=m2Kgivesonlyonesolutionm=0,n=0
Commented by 1234Hello last updated on 03/Jul/17
But why 2Kx = 4Knπ?
Butwhy2Kx=4Knπ?
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
TypoItshouldbe2mx.x=2nπforcosx=1forcos2Kx=12Kx=2mπx=2nx4Knx=2mπ

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