f(x)=根号3sin(2x-π/6)+2sin^2(x-π/12)求单调递增区间

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f(x)=根号3sin(2x-π/6)+2sin^2(x-π/12)求单调递增区间

f(x)=根号3sin(2x-π/6)+2sin^2(x-π/12)求单调递增区间
f(x)=根号3sin(2x-π/6)+2sin^2(x-π/12)求单调递增区间

f(x)=根号3sin(2x-π/6)+2sin^2(x-π/12)求单调递增区间
f(x)=√3sin(2x-π/6)+2sin²(x-π/12)
=√3sin(2x-π/6)+1-cos(2x-π/6)
=2[√3/2*sin(2x-π/6)-1/2*cos(2x-π/6)]+1
=2sin(2x-π/6-π/6)+1
=2sin(2x-π/3)+1
令2kπ-π/2≤2x-π/3≤2kπ+π/2,
得:kπ-π/12≤x≤kπ+5π/12
所以单调递增区间为[kπ-π/12,kπ+5π/12] (k∈Z)

f(x)=√3sin(2x-π/6)+2[1-cos(2x-π/6)]/2
=√3sin(2x-π/6)-cos(2x-π/6)+1
=2[sin(2x-π/6)*√3/2-cos(2x-π/6)*1/2]+1
=2[sin(2x-π/6)cosπ/6-cos(2x-π/6)sinπ/6]+1
=2sn(2x-π/3)+1
递增则2kπ-π/2<2x-π/3<2kπ+π/2
kπ-π/12(kπ-π/12,kπ+5π/12)