函數(shù)y=3sinx-cos2x的值域是
解析：∵f(x)=3sinx-cos2x
令f’(x)=3cosx+2sin2x=cosx(3+4sinx)=0
Cosx=0==>x1=2kπ-π/2,x2=2kπ+π/2
Sinx=-3/4==>x3=2kπ-arcsin(3/4),x4=(2k+1) π+arcsin(3/4)
f”(x)=-3sinx+4cos2x==>f”(x1)0
∴f(x)在,x1,x2處取極大值,f(x1)=-2,f(x2)=4
f(x)在,x3,x4處取極小值,f(x3)=f(x4)=-9/4-1+2*9/16=-17/8
∴函數(shù)y=3sinx-cos2x的值域是[-17/8,4]