函数f(x)= 2 sin(3 x +φ)+ m(φ∈R)是已知的,

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+ M = 2 sin[3π2+(3 x +φ)]+ m = f(π2+ x)= 2 sin[3(π2+ x +- (3x - φ)]+ m = -2 sin(3 x - φ)+ m ????????
?2[sin(3 x?)+ Cos(3 x +?)]+ 2 m = 2?
Sin(3x?)+ Cos(3x +?)= M?1?
Sin 3×cosφ?Cos 3×sinφ+ cos 3×cosφ?Sin 3×sinφ= m?1?
(Cosφ?Sinφ)(sin 3 x + cos 3 x)= m?1?
x∈R为真,cosφ-sinφ= 0,m-1 = 0。即,tanφ= 1,m = 1。Tanbien bitan(π - φ)= - tanφ= n,∴n= - 1。∴m + n = 0。因此,答案是0。

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