y=sin(-2x+π/6)+根号3cos(2x-π/2)的递增区间
问题描述:
y=sin(-2x+π/6)+根号3cos(2x-π/2)的递增区间
答
O(∩_∩)O~~
答
解不等式-π/2+2kπ≤-2x+π/6≤π/2+2kπ,可得增区间
-π+2kπ≤-2x-π/2≤2kπ,可得增区间
求两个解集的交集
答
y=-√3sin2x/2+cos2x/2+√3sin2x=cos(π/6)*sin2x+sin(π/6)*cos2x=sin(2x+π/6)
sinx的递增区间是[-π/2+2kπ,π/2+2kπ],所以sin(2x+π/6)的递增区间是[-π/3+2kπ,π/6+2kπ]