f(x)=根号3sin2x+2cos^2+3.若f(x)=28/5,x属于(派/6,5派/12),求cos(2x-pai

f(x)=根号3sin2x+2cos^2+3
=√3sin2x+cos2x+4
=2sin(2x+π/6)+4
若f(x)=28/5
则2sin(2x+π/6)+4=28/5
sin(2x+π/6)=4/5
已知x属于(派/6,5派/12),2x+π/6∈(π/2,π)
故cos(2x+π/6)=-√[1-sin²(2x+π/6)]=-3/5
cos(2x-π/12)=cos[(2x+π/6)-π/4]
=cos(2x+π/6)cos(π/4)+sin(2x+π/6)sin(π/4)
=(-3/5)*(√2/2)+(4/5)(√2/2)
=√2/10