Question

求函数f(x)=5√3cos平方x+√3sin平方x-4sinxcosx(∏/4≤x≤7∏/24)的最小值,并求其单调区间

请写出详细的过程，谢谢

Answer 1

f(x)=5*3^0.5*cos^2(x)+3^0.5*sin^2(x)-4*sinx*cosx (pi/4<=x<=7*pi/24) pi=3.14
=5*3^0.5*(1+cos2x)/2+3^0.5*(1-cos2x)/2-2sin2x
=3*3^0.5+2*3^0.5*cos2x-2sin2x
=3*3^0.5-4*sin(2x-pi/3)

递增区间： .2k*pi-pi/2<=2x-pi/3<=2k*pi+pi/乱橡2 =>k*pi-pi/12<拍茄=x<=k*pi+5*pi/12
递减区间： .2k*pi+pi/2<=2x-pi/3<=2k*pi+3pi/2 =>k*pi+5*pi/12<=x<袭陪察=k*pi+11*pi/12