函数f(x)=2sin(2x+π⼀6)+m+1.求f(x)在【0,π】上的单调增区间

2025-02-28 07:00:03
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回答1:

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。