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f(x）=根号3/2*sinwx +1/2*coswx +1
= sin(wx+π/6)+1
T=2π/w =π ， 所以, w=2
f(x)= sin(2x+π/6)+1
-π/2 +2kπ <= 2x +π/6 <=π/2 +2kπ
解得， -π/3 +kπ <=x<= π/6 +kπ， k∈Z
递增区间为 [ -π/3 +kπ , π/6 +kπ]， k∈Z
= sin(wx+π/6)+1
T=2π/w =π ， 所以, w=2
f(x)= sin(2x+π/6)+1
-π/2 +2kπ <= 2x +π/6 <=π/2 +2kπ
解得， -π/3 +kπ <=x<= π/6 +kπ， k∈Z
递增区间为 [ -π/3 +kπ , π/6 +kπ]， k∈Z
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