kangyingxi
幼苗
共回答了17个问题采纳率:94.1% 举报
y=-sin2x
y=-sin2x的单调递增区间为f(x)=sin2x的单调递减区间
所以,2kπ+π/2≤2x≤2kπ+3π/2
解得,kπ+π/4≤x≤kπ+3π/4(k为整数)
所以,
y=-sin2x 的单调递增区间为[kπ+π/4,kπ+3π/4](k为整数)
y=sin(π/3-2x)
y单调递增时
2kπ-π/2≤π/3-2x≤2kπ+π/2
即,2kπ-5π/6≤-2x≤2kπ+π/6
即,2kπ-π/6≤2x≤2kπ+5π/6
即,kπ-π/12≤x≤kπ+5π/12
所以,
y=sin(π/3-2x)的单调递增区间为[kπ-π/12,kπ+5π/12](k为整数)
1年前
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