cosθ+cos2θ+cos3θ=0,求θo<θ<90°
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![cosθ+cos2θ+cos3θ=0,求θo<θ<90°](/uploads/image/z/5547460-4-0.jpg?t=cos%CE%B8%EF%BC%8Bcos2%CE%B8%EF%BC%8Bcos3%CE%B8%3D0%2C%E6%B1%82%CE%B8o%EF%BC%9C%CE%B8%EF%BC%9C90%C2%B0)
cosθ+cos2θ+cos3θ=0,求θo<θ<90°
cosθ+cos2θ+cos3θ=0,求θ
o<θ<90°
cosθ+cos2θ+cos3θ=0,求θo<θ<90°
原式=cos(x+2x)=cosxcos2x-sinxsin2x=cosx((cosx)^2-(sinx)^2)-2(sinx)^2cosx
=(cosx)^3-3cosx(sinx)^2
=cosx(1-4(sinx)^2)
cosx=0 或者(sinx)^2=1/4
x=(k+1/2)pi 或者x=(k/2+1/4)pi k属于Z