f(x)=5√3cos²x+√3sin²x-4sinxcosx
=4√3cos²x+√3-4sinxcosx
=2√3(2cos²x-1)+3√3-2*2sinxcosx
=2√3cos2x-2sin2x+3√3
=4(√3/2*cos2x-1/2*sin2x)+3√3
=4(cos2x*cosπ/6-sin2x*sinπ/6)+3√3
=4cos(2x+π/6)+3√3.
因为π/4≤x≤7π/24,
所以2π/3≤2x+π/6≤3π/4,
-√2/2