函数y=2sinx(sinx+cosx)的单调递减区间是?
网友回答
y=2sinx(sinx+cosx)=2Sin²x+2SinxCosx
=1-Cos2x+Sin2x
=√2Sin(2x-π/4)+1
单调递减则2x-π/4∈[2kπ+π/2,2kπ+3π/2],k∈Z
x∈[kπ+3π/8,kπ+7π/8],k∈Z
函数y=2sinx(sinx+cosx)的单调递减区间是[kπ+3π/8,kπ+7π/8],k∈Z
======以下答案可供参考======
供参考答案1:
y=2(sinx)^2+2sinxcosx
y的导数=4sinxcosx+2cos2x=2sin2x+2cos2x=根号2sin(2x+π/4)
所以只要sin(2x+π/4)2kπ+π供参考答案2:
求导 找零点 划区间