Question

f(x)=sin(-2x+兀/6)+1,求f(x)的单调区间

Answer 1

问：f(x)=sin(-2x+兀/6)+1,求f(x)的单调区间

答：单调递增区间-π/2+2kπ＜-2x+π＜π/2+2kπ-3π/2+2kπ＜-2x＜-π/2+2kππ/4+kπ＜x＜3π/4+kπ

答：单调递减区间π/2+2kπ＜-2x+π＜3π/2+2kπ-π/2+2kπ＜-2x＜π/2+2kπ-π/4+kπ＜x＜π/4+kπ

问：怎么求的

答：上面那个就是详细步骤呀

答：哪里没看懂吗

问：单调递减区间π/2+2kπ≤-2x+π/6≤3π/2+2kπ怎么算

答：这是sin函数单调递减区间带进去呀

问：f（x）=sin（-2x 兀/6） 1，当x取何值时，函数有最大值？