x^2cosC+4xsin+6?sin?sinc根据题意，x^2cosC+4xsinC+6>0对一切实数x恒成立,则cosC<>0,于是x^2cosC+4xsinC+6=cosC(x+2sinC/cosC)^2+6-4sin^2C/cosC>0若1>cosC>0,则(x+2sinC/cosC)^2>4sin^2C/cos^2C-6/cosC=4/cos^2C-6/cosC-4=4(1/cosC-3/4)^2-25/4所以4(1/cosC-3/4)^2-25/4<0,-5/4<1/cosC-3/4<5/4,-1/2<1/cosC<2所以1>=cosC>1/2,(舍去cosC<-2)若-10,1/cosC-3/4<-5/4,cosC>-2,(舍去cosC>1/2)所以-1=cosC>1/2