若函数f(x)=-4sin^2x+4√3sinxcosx+m-2,当x属于(0,90°]时f(x)的值域为[-5,1],求m的值.(以上的是平方和4倍根号3)
热心网友
f(x)=-4(sinx)^2+4√3sinxcosx+m-2=2-4(sinx)^2+4√3sinxcosx+m-4=2cos2x+2√3sin2x+m-4=4[(1/2)cos2x+(√3/2)sin2x]+m-4=4sin[2x+(π/6)]+m-4因x∈(0,π/2] == 2x∈(0,π] == 2x+(π/6)∈(π/6,7π/6]所以2x+(π/6)]=7π/6时，fmin=m-62x+(π/6)]=π/2时，fmax=m既f(x)的值域为[m-6，m]由已知f(x)的值域为[-5,1],==m-6=-5, m=1
热心网友
f(x)=-4sin^2x+4√3sinxcosx+m-2=-2(1-cos2x)+2√3sin2x+m-2=2(cos2x+√3sin2x)+m-4=4sin(2x+30°)+m-430°<2x+30°<=210°-2=<4sin(x+30°)<=4所以m-4-2=-5m=1