y=sin(π/3-x/2)=-sin(x/2-π/3)
令 2kπ-π/2≤x/2-π/3≤2kπ+π/2,k∈Z
解得4kπ-π/3≤x≤4kπ+5π/3,k∈Z
又∵x∈[-2π,2π]
∴當(dāng)k=0時,x∈[-π/3,5π/3]
即函數(shù)的單調(diào)遞減區(qū)間是[-π/3,5π/3].題目是y=2sin(π/3-χ/2)呀 在sin前有個2啊一樣的，前面的系數(shù)不影響單調(diào)區(qū)間，放心采納吧，呵呵。 y=2sin(π/3-x/2)=-2sin(x/2-π/3)令 2kπ-π/2≤x/2-π/3≤2kπ+π/2，k∈Z解得4kπ-π/3≤x≤4kπ+5π/3，k∈Z又∵x∈[-2π，2π]∴當(dāng)k=0時，x∈[-π/3，5π/3]即函數(shù)的單調(diào)遞減區(qū)間是[-π/3，5π/3].