已知f(x)=2sin(2x+π/6),若f(x0)=6/5,x0∈[π/4,π/2],则cos2x0=?

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已知f(x)=2sin(2x+π/6),若f(x0)=6/5,x0∈[π/4,π/2],则cos2x0=?已知f(x)=2sin(2x+π/6),若f(x0)=6/5,x0∈[π/4,π/2],则co

已知f(x)=2sin(2x+π/6),若f(x0)=6/5,x0∈[π/4,π/2],则cos2x0=?
已知f(x)=2sin(2x+π/6),若f(x0)=6/5,x0∈[π/4,π/2],则cos2x0=?

已知f(x)=2sin(2x+π/6),若f(x0)=6/5,x0∈[π/4,π/2],则cos2x0=?
x0∈[π/4,π/2];
=> (1)2x0∈[π/2,π];
(2)2x0+π/6∈[2π/3,7π/6];
将X0代入,f(x0)=2sin(2x0+π/6)=6/5
=> sin(2x0+π/6)=3/5;
由（2）得 cos(2x0+π/6)=-4/5
=> cos2x0=sin(2x0+π/2)=sin[(2x0+π/6)+π/3]=sin(2x0+π/6)cosπ/3+cos(2x0+π/6)sinπ/3
=3/5*1/2+(-4/5)*(√3)/2=3/10-(4√3)/10=(3-4√3)/10
打的累死了

0.9797

如图，看得清么？

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