已知函数f(x)=1/2sin²x+cos²x+根号3/2sinxcosx 求函数最小正周期 求出函数的单调区间
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f(x)=1/2sin²x+cos²x+根号3/2sinxcosx=1/2+1/2cos^2 x+√3/2sinxcosx
=1/2+1/4(1+cos2x)+√3/4sin2x
=3/4+1/2sin(2x+π/6)
T=2π/2=π
令2kπ-π/2<=2x+π/6<=2kπ+π/2
得到kπ-π/3<=x<=kπ+π/6
所以函数的单调递增区间是[kπ-π/3,kπ+π/6] (k是整数)
令2kπ+π/2<=2x+π/6<=2kπ+3π/2
得到kπ+π/6<=x<=kπ+2π/3
所以函数的单调递减区间是[kπ+π/6，kπ+2π/3] (k是整数)
=1/2+1/4(1+cos2x)+√3/4sin2x
=3/4+1/2sin(2x+π/6)
T=2π/2=π
令2kπ-π/2<=2x+π/6<=2kπ+π/2
得到kπ-π/3<=x<=kπ+π/6
所以函数的单调递增区间是[kπ-π/3,kπ+π/6] (k是整数)
令2kπ+π/2<=2x+π/6<=2kπ+3π/2
得到kπ+π/6<=x<=kπ+2π/3
所以函数的单调递减区间是[kπ+π/6，kπ+2π/3] (k是整数)
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