已知函数f(x)=sin^x+2√3sin(x+π/4)cos(x-π/4)-cos^2x-√3

求函数的最小正周期和单调递减区间

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解：已知函数的表达式为：

f(x) = sin²x + 2√3sin(x+π/4)cos(x-π/4) -- cos²x -- √3 对中运吧？

细节分析：
① cos2x = cos(x + x)
= cosx cosx -- sinx sinx
= cos²x -- sin²x

② sin(x + π/4) = sinx cos(π/4) + cosx sin(π/4)
= (√2/2) sinx + (√2/2) cosx
= (√2/2) (sinx + cosx)

③ cos(x -- π/4) = cosx cos(π/4) + sinx sin(π/4)
= (√2/2) cosx + (√2/2) sinx
= (√2/2) (sinx + cosx)

把以上三式代入原函数表达式，得：
f(x) = sin²x + 2√3sin(x+π/4)cos(x-π/4) -- cos²x -- √3

= sin²x -- cos²x + 2√3sin(x+π/4)cos(x-π/4) -- √3

= -- (cos²x -- sin²x) + 2√3sin(x+π/4)cos(x-π/4) -- √3

= -- cos2x + 2√3 × [(√2/2) (sinx + cosx)]² -- √3

= -- cos2x + 2√3 × [ (1/2) × (sinx + cosx)² ] -- √3

= -- cos2x + √3 × (sinx + cosx)² -- √3

= -- cos2x + √3 × (1 + 2 sinx cosx) -- √3 (注：sin²x + cos²x = 1)

= -- cos2x + √3 × (1 + sin2x) -- √3 (注：sin2x = 2 sinx cosx)

= √3 sin2x -- cos2x

= 2 × [ (√3/2) sin2x -- (1/2) cos2x ]

= 2 × [ sin2x cos(π/6) -- cos2x sin(π/6) ]

= 2sin(2x -- π/6)

∴ 其最小正周期为：
T = 2π/ 2
= π

再求其单调递减区间：

设X = 2x -- π/6，

而运团sinX的单调递减区间为 [ 2kπ + π/2，2kπ + 3π/卖悄梁2 ]

∴ 2kπ + π/2 ≤ 2x -- π/6 ≤ 2kπ + 3π/2

∴2kπ + π/2 + π/6 ≤ 2x ≤ 2kπ + 3π/2 + π/6

∴2kπ + 2π/3 ≤ 2x ≤ 2kπ + 5π/3

∴ kπ + π/3 ≤ x ≤ kπ + 5π/6

∴ f（x）的单调递减区间为 [ kπ + π/3，kπ + 5π/6 ] （k ∈Z）。

祝您学习顺利！

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问题不详，无法解答

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