what is the value of x at infinite
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y=sin(x)sin(2x), dy/dx=sin(2x)cos(x)+2sin(x)cos(2x)=0 at extrema.

sin(2x)=2sin(x)cos(x), so y=2sin2(x)cos(x).

Expanding: 2sin(x)(1-sin2(x))+2sin(x)(1-2sin2(x))=0,

2sin(x)(2-3sin2(x))=0. x=0 is a solution but this appears to be a minimum because y=0, when x=0.

sin2(x)=⅔, cos2(x)=1-⅔=⅓, making cos(x)=1/√3 or √3/3. y=2sin2(x)cos(x)=2×⅔×√3/3=4√3/9=0.7698 approx.

by Top Rated User (1.1m points)

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