已知函数f(x)=4cossin(x+π/6)-1求函数单调递减区间
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f(x)=4cosxsin(x+π/6)-1
=4cosx(sinxcosπ/6+cosxsinπ/6)-1
=4×√3/2sinxcosx+4×1/2×cos²x-1
=√3sin2x+cos2x
=2(√3/2sin2x+1/2cos2x)
=2sin(2x+π/6)
由2kπ+π/2≤2x+π/6≤2kπ+3π/2
得 kπ+π/6≤x≤kπ+2π/3，k∈Z
∴f(x)单调递减区间为 [kπ+π/6,kπ+2π/3]，k∈Z
f(x)=4cosxsin(x+π/6)-1
=4cosx(sinxcosπ/6+cosxsinπ/6)-1
=4×√3/2sinxcosx+4×1/2×cos²x-1
=√3sin2x+cos2x
=2(√3/2sin2x+1/2cos2x)
=2sin(2x+π/6)
由2kπ+π/2≤2x+π/6≤2kπ+3π/2
得 kπ+π/6≤x≤kπ+2π/3，k∈Z
∴f(x)单调递减区间为 [kπ+π/6,kπ+2π/3]，k∈Z
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