①sin x-((√3)/2)≥0
sin x≥√3)/2;
x∈[2kπ+π/3,2kπ+2π/3];k∈Z;
②√2-(2cos x)＞0;
cos x＜√2/2;
x∈(2kπ+π/4,2kπ+7π/4);k∈Z
∴x∈[2kπ+π/3,2kπ+2π/3];k∈Z
求函数y=(√[sin x-((√3)/2)])+lg[√2-(2cos x)]的定义域
①sin x-((√3)/2)≥0
sin x≥√3)/2;
x∈[2kπ+π/3,2kπ+2π/3];k∈Z;
②√2-(2cos x)＞0;
cos x＜√2/2;
x∈(2kπ+π/4,2kπ+7π/4);k∈Z
∴x∈[2kπ+π/3,2kπ+2π/3];k∈Z