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2sin2x 3sinx 0

2sin²x-(2a+3)sinx+(4a-2)=0 有实根 ①(2a+3)²-8(4a-2)>0→4a²+25>0恒成立 a为任意值 ②-1≤sinx=[(2a+3)±√4a²+25]/4≤1 令f(a)=[(2a+3)±√4a²+25]/4 f'(a)=(2[2±1/2√4a²+25)/4=1±1/4√4a²+25 取正号时,f'(a)恒大...

可以的啊 这一题

2sinx 与 sin2x 同阶, 同阶的两个函数相减,再用同阶的函数代替,不就减完了么?? 只有在它们相除时,可以用同阶的代替 。

f(x)=2cosx(3sinx?cosx)+1=2sin(2x-π6)∴要得到函数f(x)=2sin(2x-π6),需将函数y=2sin2x的图象向右平移π12k(k∈N)个单位,∴φ的最小值为π12故答案为:π12

f(x)=√3sin2x+2sin²x =√3sin2x-(1-2sin²x)+1 =√3sin2x-cos2x+1 =2[sin2xcos(π/6)-cos2xsin(π/6)]+1 =2sin(2x-π/6)+1. ∵-1≤sin(2x-π/6)≤1, ∴sin(2x-π/6)=-1时,f(x)|min=-1. sin(2x-π/6)=1时,f(x)|max=3。

题目1:(1)f(θ)=-4,2sin(2θ-π/3)-3=-4,sin(2θ-π/3)=-1/2 θ∈[0,2π],2θ-π/3∈[-π/3,11π/3] 得2θ-π/3=-π/6,7π/6,2π-π/6,2π+7π/6 所以θ=π/12,3π/4,13π/12,7π/4 (2)|f(x)-m|

f(x) = 4sinx(cosx-sinx)+3 = 4sinxcosx-4sin²x+3 = 2sin2x-2(1-cos2x)+3 = 2sin2x+2cos2x+1 = 2√2(sin2xcosπ/4+cos2xsinπ/4) + 1 = 2√2sin(2x+π/4) + 1 x∈(0,π) 2x∈(0,2π) 2x+π/4∈(π/4,9π/4) 2x+π/4∈(π/2,3π/2)时单调递减 此...

∵f(x)=sin2x-2sinxcosx+3cos2x=1-2sinxcosx+2cos2x=1+1+cos2x-sin2x=-(sin2x-cos2x)+2=-2sin(2x-π4)+2.∴把函数f(x)的图象沿x轴向左平移m(m>0)个单位,得到函数g(x)的图象的解析式为:g(x)=-2sin(2x+2m-π4)+2.∵函数g(x)的图象关...

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