1
f(x)=coc(2x+π/3)-1/2cos2x+1/2
= 1/2*cos2x-√3/2*sin2x-1/2cos2x+1/2
=-√3/2*sin2x+1/2
当2x=2kπ-π/2（k∈Z）时,f(x)max=(√3+1)/2
最小正周期T=π
2
f(C/3)=-1/4
-√3/2*sin(2C/3)+1/2=-1/4
==>sin(2C/3)=√3/2
∵0
设函数f(x)=coc(2x+π/3)-1/2cos2x+1/2
3个回答
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