arcsin(x-3)-π/6≥0arcsin(x-3)≥π/6x-3≥1/2x≥7/2而-1≤x-3≤1,即2≤x≤4所以7/2≤x≤4,即定义域为[7/2,4]而因为π/6≤arcsin(x-3)≤π/2所以0≤arcsin(x-3)-π/6≤π/3所以0≤√[arcsin(x-3)-π/6]≤√(3π)/3...