已知函数f(x)=根号3sinxcosx+cos的平方x+1，求f(x)的最小正周期及单调递减区间
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(1)f(x)=√3sinxcosx+cos�0�5x
=√3/2sin2x+(cos2x+1)/2
=√3/2sin2x+1/2cos2x+1/2
=sin(2x+π/6)+1/2
f(x)的最小正周期为2π/2=π
2kπ-π/2≤2x+π/6≤2kπ+π/2 (k∈Z)
kπ-π/3≤x≤kπ+π/6
则f(x)的单调递增区间为[kπ-π/3,kπ+π/6] (k∈Z)
2kπ≤2x+π/6≤2kπ+π (k∈Z)单调递减区间[kπ-π/12,kπ+5π/12]
=√3/2sin2x+(cos2x+1)/2
=√3/2sin2x+1/2cos2x+1/2
=sin(2x+π/6)+1/2
f(x)的最小正周期为2π/2=π
2kπ-π/2≤2x+π/6≤2kπ+π/2 (k∈Z)
kπ-π/3≤x≤kπ+π/6
则f(x)的单调递增区间为[kπ-π/3,kπ+π/6] (k∈Z)
2kπ≤2x+π/6≤2kπ+π (k∈Z)单调递减区间[kπ-π/12,kπ+5π/12]
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