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Question-88029




Question Number 88029 by M±th+et£s last updated on 07/Apr/20
Answered by mind is power last updated on 08/Apr/20
−4sin^2 (4x)−8sin(4x)−9cos^2 (4x)+12cos(4x)−4+6sin(8x)  =−(2sin(4x)−3cos(4x)+2)^2   f(x)=((−(2sin(4x)−3cos(4x)+2)^2 +1+4−6cos(4x)+4sin(4x))/(2sin(4x)−3cos(4x)+2))  let Y(x)=2sin(4x)−3cos(4x)+2  g(y)=−Y+(1/Y)+2  g′(y)=−1−(1/y^2 )  y(x+(π/2))=y(x)  2sin(4x)−3cos(4x)+2=0  ⇒(√(13))sin(4x−arcsin((2/( (√(13))))))+2=0  has solution  min and max didnt existe
4sin2(4x)8sin(4x)9cos2(4x)+12cos(4x)4+6sin(8x)=(2sin(4x)3cos(4x)+2)2f(x)=(2sin(4x)3cos(4x)+2)2+1+46cos(4x)+4sin(4x)2sin(4x)3cos(4x)+2letY(x)=2sin(4x)3cos(4x)+2g(y)=Y+1Y+2g(y)=11y2y(x+π2)=y(x)2sin(4x)3cos(4x)+2=013sin(4xarcsin(213))+2=0hassolutionminandmaxdidntexiste
Commented by M±th+et£s last updated on 08/Apr/20
god bless you sir
godblessyousir

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