过程:
2sin(2x-π/4)≥√3
sin(2x-π/4)≥√3/2
sin(2x-π/4)≥sinπ/3
作图,在[0,2π]之中sinπ/3所在的与x轴平行的直线与图像有两个交点,分别对应π/3,2π/3,所以此两点构成的区域即为2x-π/4的取值范围,得:
π/3≤2x-π/4≤2π/3
7π/24≤x≤11π/24
解不等式:2sin(2x-π/4)≥√3
解:sin(2x-π/4)≥√3/2
∴2kπ+π/3≤2x-π/4≤2kπ+π/2
2kπ+π/3+π/4≤2x≤2kπ+π/2+π/4
2kπ+7π/12≤2x≤2kπ+3π/4
∴ kπ+7π/24≤x≤kπ+3π/8 (k∈Z).