cos(2x+π/3)+sin^2x-1/2 求x在 0到π之间时,函数单调递减区间
问题描述：
cos(2x+π/3)+sin^2x-1/2 求x在 0到π之间时,函数单调递减区间
答
y=cos(2x+π/3)+sin²x -1/2
=cos2xcos(π/3)-sin2xsin(π/3)+(1-cos2x)/2 -1/2
=-(√3/2)sin2x
令 -π/2+2kπ≤2x≤π/2+2kπ,
解得 -π/4+kπ≤x≤π/4+kπ,k 是整数
因为 0≤x≤π,从而 k取0,1,得函数单调递减区间为[0,π/4]和[3π/4,π]第二部怎么得出了第三部=cos2xcos(π/3)-sin2xsin(π/3)+(1-cos2x)/2 -1/2=cos2x·(1/2) -sin2x·(√3/2)+1/2 -cos2x·(1/2) -1/2=-(√3/2)sin2x