f(x)=cos2x*1/2-√3*sin2x+(1-cos2x)/2
=cos2x-√3sin2x+1/2
=2cos(2x+π/3)+1/2
所以最小正周期T=2π/2=π
当cos（2x+π/3）=1
取最大值
为5/2
设函数f(x)=cos(2x+π/3)+sin^2 X
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