求函数y=2sinx(1-sinx)\3-cos2x+4sinx,x属于(0,π\2)的值域

求函数y=2sinx(1-sinx)\3-cos2x+4sinx,x属于(0,π\2)的值域

y=2sinx(1-sinx)/3-cos2x+4sinx
=(2/3)sinx-(2/3)(sinx)^2-[1-2(sinx)^2]+4sinx
=(2/3)sinx-(2/3)(sinx)^2-1+2(sinx)^2+4sinx
=(4/3)(sinx)^2+(14/3)sinx-1
令sinx=t,则0y=4t^2/3+14t/3-1
=(4/3)(t+7/4)^2-61/12
所以-1