已知,函数f(x)=2sin(2x+π/6)+1求它在[0,π]上的单调递增.f(x)=2sin(2x+π/6)+1记t=2x+π/6,t∈[π/6,2π+π/6]f(t)=2sin(t)+1f(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]

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