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/12)的值

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=28/5sin(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)]=-3/5cos(2x-...