已知函数f(x)=sinxsin(x+π/3)-1

解 和差化积,可得 2f(x)+2=2sinxsin[x+60º]=cos60º-cos[2x+60º]=(1/2)-cos[2x+60º] ∴f(x)=(-3/4)-[cos(2x+60º)/2] [[[1]]] f(x)≥-1 可得:cos(2x+60º)≤1/2. 2kπ+(π/3)≤2x+(π/3)≤2kπ+(2π-π/3) kπ≤x≤kπ+(2π/3) ∴