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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

$$-4{sin}^2\left(4x\right)-8sin\left(4x\right)-9{cos}^2\left(4x\right)+12cos\left(4x\right)-4+6sin\left(8x\right)$$$$=-{(2sin\left(4x\right)-3cos\left(4x\right)+2)}^2$$$$f\left(x\right)=\frac{-{(2sin\left(4x\right)-3cos\left(4x\right)+2)}^2+1+4-6cos\left(4x\right)+4sin\left(4x\right)}{2sin\left(4x\right)-3cos\left(4x\right)+2}$$$$letY\left(x\right)=2sin\left(4x\right)-3cos\left(4x\right)+2$$$$g\left(y\right)=-Y+\frac{1}{Y}+2$$$$g^{\prime }\left(y\right)=-1-\frac{1}{y^2}$$$$y(x+\frac{\pi }{2})=y\left(x\right)$$$$2sin\left(4x\right)-3cos\left(4x\right)+2=0$$$$\Rightarrow \sqrt{13}sin(4x-arcsin\left(\frac{2}{\sqrt{13}}\right))+2=0$$$$hassolutionminandmaxdidntexiste$$$$$$$$$$

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

$$godblessyousir$$

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