已知函数f（x）＝2根号3 sinxcosx＋2cos^2x-1(x属于R) 若f（x）＝6/5，x属于〔π/4，π/2〕,求cos2x的值
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f(x)=2倍根号3sinxcosx+2cos^2x-1
=√3sin2x+cos2x
=2sin(2x+π/6)
2sin(2x0+π/6)=6/5
sin(2x0+π/6)=3/5 cos(2x0+π/6)=-4/5
cos2xo=cos[(2x0+π/6)-π/6]
=cos(2x0+π/6)cosπ/6+sin(2x0+π/6)sinπ/6
=-4/5*√3/2+3/5*1/2
=(3-4√3)/10
=√3sin2x+cos2x
=2sin(2x+π/6)
2sin(2x0+π/6)=6/5
sin(2x0+π/6)=3/5 cos(2x0+π/6)=-4/5
cos2xo=cos[(2x0+π/6)-π/6]
=cos(2x0+π/6)cosπ/6+sin(2x0+π/6)sinπ/6
=-4/5*√3/2+3/5*1/2
=(3-4√3)/10
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