函数f(x)=2sin(2x+π/6)+m+1.求f(x)在【0,π】上的单调增区间

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函数f(x)=2sin(2x+π/6)+m+1.求f(x)在【0,π】上的单调增区间

函数f(x)=2sin(2x+π/6)+m+1.求f(x)在【0,π】上的单调增区间
函数f(x)=2sin(2x+π/6)+m+1.求f(x)在【0,π】上的单调增区间

函数f(x)=2sin(2x+π/6)+m+1.求f(x)在【0,π】上的单调增区间
f(x)=2sin(2x+π/6)+m+1
记t=2x+π/6,t∈[π/6,2π+π/6]
f(t)=2sin(t)+m+1
f(t)的单调递增区间为：
t∈[π/6,π/2]或t∈[3π/2,2π+π/6]
此时：
x∈[0,π/6]或x∈[2π/3,π]
即所求.

2kπ-π/2<2x+π/6<2kπ+π/2
kπ-π/32kπ+π/2<2x+π/6<2kπ+3π/2
kπ+π/6kπ-π/3（0，π/6）递增
kπ+π/6（π/6，π）递减

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