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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

$$−\mathrm{4}{sin}^{\mathrm{2}} \left(\mathrm{4}{x}\right)−\mathrm{8}{sin}\left(\mathrm{4}{x}\right)−\mathrm{9}{cos}^{\mathrm{2}} \left(\mathrm{4}{x}\right)+\mathrm{12}{cos}\left(\mathrm{4}{x}\right)−\mathrm{4}+\mathrm{6}{sin}\left(\mathrm{8}{x}\right) \\ $$$$=−\left(\mathrm{2}{sin}\left(\mathrm{4}{x}\right)−\mathrm{3}{cos}\left(\mathrm{4}{x}\right)+\mathrm{2}\right)^{\mathrm{2}} \\ $$$${f}\left({x}\right)=\frac{−\left(\mathrm{2}{sin}\left(\mathrm{4}{x}\right)−\mathrm{3}{cos}\left(\mathrm{4}{x}\right)+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{1}+\mathrm{4}−\mathrm{6}{cos}\left(\mathrm{4}{x}\right)+\mathrm{4}{sin}\left(\mathrm{4}{x}\right)}{\mathrm{2}{sin}\left(\mathrm{4}{x}\right)−\mathrm{3}{cos}\left(\mathrm{4}{x}\right)+\mathrm{2}} \\ $$$${let}\:{Y}\left({x}\right)=\mathrm{2}{sin}\left(\mathrm{4}{x}\right)−\mathrm{3}{cos}\left(\mathrm{4}{x}\right)+\mathrm{2} \\ $$$${g}\left({y}\right)=−{Y}+\frac{\mathrm{1}}{{Y}}+\mathrm{2} \\ $$$${g}'\left({y}\right)=−\mathrm{1}−\frac{\mathrm{1}}{{y}^{\mathrm{2}} } \\ $$$${y}\left({x}+\frac{\pi}{\mathrm{2}}\right)={y}\left({x}\right) \\ $$$$\mathrm{2}{sin}\left(\mathrm{4}{x}\right)−\mathrm{3}{cos}\left(\mathrm{4}{x}\right)+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\sqrt{\mathrm{13}}{sin}\left(\mathrm{4}{x}−{arcsin}\left(\frac{\mathrm{2}}{\sqrt{\mathrm{13}}}\right)\right)+\mathrm{2}=\mathrm{0} \\ $$$${has}\:{solution}\:\:{min}\:{and}\:{max}\:{didnt}\:{existe} \\ $$$$ \\ $$$$ \\ $$

Commented by M±th+et£s last updated on 08/Apr/20

god bless you sir

$${god}\:{bless}\:{you}\:{sir} \\ $$

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