已知函数fx=2sinxcos(x＋π÷3)＋根号3÷2求函数单调递减区间
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f(x)=2sinxcos(x＋π/3)＋√3/2
=2sinx(1/2cosx-√3/2sinx)＋√3/2
=1/2*2cosx-√3/2*2(sinx)^2＋√3/2
=1/2sin2x-√3/2(1-cos2x)＋√3/2
=1/2sin2x+√3/2cos2x
=sin(2x+π/3)
因:2x+π/3∈[2kπ+π/2,2kπ+3π/2]是单调递减
所以:x∈[kπ+π/12,kπ+7π/12]是单调递减
=2sinx(1/2cosx-√3/2sinx)＋√3/2
=1/2*2cosx-√3/2*2(sinx)^2＋√3/2
=1/2sin2x-√3/2(1-cos2x)＋√3/2
=1/2sin2x+√3/2cos2x
=sin(2x+π/3)
因:2x+π/3∈[2kπ+π/2,2kπ+3π/2]是单调递减
所以:x∈[kπ+π/12,kπ+7π/12]是单调递减
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