y=sin(兀/3-1/2x),x∈[-2兀,2兀]增区间?
问题描述：
y=sin(兀/3-1/2x),x∈[-2兀,2兀]增区间?
答
y=sin(π/3-x/2)=-sin(x/2-π/3)
增区间,即y=sin(x/2-π/3)的减区间
2kπ+π/2≤x/2-π/3≤2kπ+3π/2
2kπ+5π/6≤x/2≤2kπ+11π/6
即 4kπ+5π/3≤x≤4kπ+11π/3
结合题目中的x∈[-2π,2π]
∴ 增区间是[5π/3,2π] 和[-2π,-π/3]