求fx=4cosxsin(x+π/6)-1求此函数的单调区间

paopaoxing2022-10-04 11:39:541条回答

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dandan658636 共回答了19个问题 | 采纳率100%
fx=4cosxsin(x+π/6)-1
=4cosx[sinx(√3/2)+(1/2)cosx]-1
=2√3sinxcosx+2cos²x-1
=√3sin2x+cos2x
=2sin(2x+π/6)
单调增区间:
2x+π/6∈[2kπ-π/2,2kπ+π/2]
x∈[kπ-π/3,kπ+π/6]
单调增区间为 [kπ-π/3,kπ+π/6]
单调减区间为 [kπ+π/6,kπ+2π/3]
1年前

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fx=4cosxsin(x+π/6)-1=4cosx[sinx(√3/2)+(1/2)cosx]-1=2√3sinxcosx+2cos²x-1=√3sin2x+cos2x=2sin(2x+π/6)单调增区间:2x+π/6∈[2kπ-π/2,2kπ+π/2]x∈[kπ-π/3,kπ+π/6]单调增区间为 [kπ-π/3,kπ+...