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Question Number 177933 by Spillover last updated on 11/Oct/22

If 1+sin x+sin^2 x+sin^3 x+...∞  =4+2(√3)  ,0<x<π Find the value of x

If1+sinx+sin2x+sin3x+... =4+23,0<x<πFindthevalueofx

Answered by Ar Brandon last updated on 11/Oct/22

1+sinx+sin^2 x+sin^3 x+∙∙∙=(1/(1−sinx))=4+2(√3)  ⇒(1/(1−sinx))=2(2+(√3))=(2/(2−(√3)))=(1/(1−((√3)/2)))  ⇒sinx=((√3)/2) ⇒x=(π/3), ((2π)/3)

1+sinx+sin2x+sin3x+=11sinx=4+23 11sinx=2(2+3)=223=1132 sinx=32x=π3,2π3

Commented bySpillover last updated on 11/Oct/22

thanks for solving

thanksforsolving

Answered by Rasheed.Sindhi last updated on 11/Oct/22

1+sin x+sin^2 x+sin^3 x+...∞=4+2(√3)  sin x+sin^2 x+sin^3 x+...∞=3+2(√3)  sin x(1+sin x+sin^2 x+sin^3 x+...∞)=3+2(√3)  sin x(4+2(√3))=3+2(√3)   sin x=((3+2(√3))/(4+2(√3)))∙((4−2(√3))/(4−2(√3)))=((12−4(3)+8(√3) −6(√3))/(16−4(3)))  sin x=((2(√3))/4)=((√3)/2)⇒x=60°,120°

1+sinx+sin2x+sin3x+...=4+23 sinx+sin2x+sin3x+...=3+23 sinx(1+sinx+sin2x+sin3x+...)=3+23 sinx(4+23)=3+23 sinx=3+234+23423423=124(3)+8363164(3) sinx=234=32x=60°,120°

Commented bySpillover last updated on 11/Oct/22

nice solution

nicesolution

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