已知f(x)=2sin(x+m/2)cos(x+m/2)+2√3[sin(x+m/2)]^2-√31.化简f(x)解析式2.若0≤m≤2П，且f(x)为偶函数，求m3.在2的条件下求满足f(x)=1,x∈[-П.П]的x的集合
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热心网友
1。化简f(x)解析式f(x)=2sin(x+m/2)cos(x+m/2)+2√3[sin(x+m/2)]^2-√3=sin(2x+m)+(√3)[1-cos(2x+m)]-√3=sin(2x+m)-(√3)cos(2x+m)=2sin[2x+m-(π/3)]2。若0≤m≤2π，且f(x)为偶函数，求m∵f(x)为偶函数∴f(-x)=f(x)即sin[2x+m-(π/3)]=sin[-2x+m-(π/3)]此进2x+m-(π/3)=-2x+m-(π/3)或2x+m-(π/3)+[-2x+m-(π/3)]=π+2kπ解得：x=0或m=(6k+5)π/6∵0≤m≤2π∴m=5π/6，11π/6，17π/63。在2的条件下求满足f(x)=1,x∈[-π。π]的x的集合 ∵f(-x)=f(x)=1且f(x)=2sin[2x+m-(π/3)]∴sin[2x+m-(π/3)]=1/2剩下自己想想吧。