Question

设函数f(ⅹ)=5√3cosv2x+√3ⅴ2x-4sⅰηxc0sⅹ(|)求f(ζπ/2)

求函数f(x)=5√3cos^2x+√3sin^2x-4sinxcosx(x属于[π/4,7π/24])的最小值

Answer 1

f(x)=5√3/2（cos2x+1)+√3/2(1-cos2x)-2sin2x=2√3cos2x-2sin2x+3√3=3√3-4sin(2x-π/3)当x=7π/24时f(x)最小值为3√3-2√2