Question

已知函数f(x)=根号3sin2x-2(sinx)^2 (1)若x属于【-30°,60°】,求f(X)的值域

Answer 1

f(x)=sqrt(3)sin2x-2sinx^2=sqrt(3)sin2x+cos2x-1=2sin(2x+π/6)-1

因为-π/6≤x≤π/3,所以-π/6≤2x+π/6≤5π/6

当2x+π/6=-π/6,即x=-π/6时,f(x)取得最小值：2*(-1/2)-1=-2

当2x+π/6=π/2,即x=π/6时,f(x)取得最大值：2*1-1=1

所以f(x)的值域：[-2,1]