已知函数f(x)=2√3sinxcosx+2(cosx)平方-1,g(x)=|f(x)|,求函数g(x)的单调递减区间,需要转

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2024-12-03 23:51:20
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回答(1):

f(x) = 2√3sinxcosx + 2cos^2x-1
= √3sin2x + cos2x
= 2(sin2xcosπ/6+cos2xsinπ/6)
= 2sin(2x+π/6)

g(x) = | f(x) | = 2 | sin(2x+π/6) |
x∈【kπ-7π/12,kπ-π/12】时,2x+π/6∈【2kπ-π,2kπ】,g(x) = -2 sin(2x+π/6)
x∈【kπ-π/12,kπ+5π/12】时,2x+π/6∈【2kπ,2kπ+π】,g(x) =2 sin(2x+π/6)

2x+π/6 ∈(2kπ-π/2,2kπ),(2kπ+π/2,2kπ+π)时,g(x)单调减
∴单调减区间:(kπ-π/3,kπ-π/12),(kπ+π/6,kπ+5π/12)