求函数f(x)=sin(π/3+4x)+cos(4x-π/6)的最小正周期和递减区间
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求函数f(x)=sin(π/3+4x)+cos(4x-π/6)的最小正周期和递减区间
求函数f(x)=sin(π/3+4x)+cos(4x-π/6)的最小正周期和递减区间
求函数f(x)=sin(π/3+4x)+cos(4x-π/6)的最小正周期和递减区间
sin(π/3+4x)=cos[π/2-(π/3+4x)]=cos(π/6-4x)
f(x)=sin(π/3+4x)+cos(4x-π/6)=2cos(4x-π/6)
最小正周期是:T=2π/4=π/2
2kπ=<4x-π/6<=2kπ+π
2kπ+π/6=<4x<=2kπ+7π/6
(kπ/2)+(π/24)=
sin(PI/3+4X)=cos((PI/2)-(PI/3+4X))=cos(-4X+PI/6)=cos(4X-PI/6);在自己算!