Question

已知f(x)=2sin^2x+2sinxcosx,(1)求f(x)的最小正周期及单调递减区间

Answer 1

解：f(x)=2sin^2x+2sinxcosx
=1-cos2x+sin2x
=1+根号2*sin(2x- π/4)
则可知f(x)的最小值周期T=2π/2=π
而当π/2 +2kπ≤2x- π/4≤3π/2 +2kπ即3π/8 +kπ≤2x- π/4≤7π/8 +kπ，k属于Z时，函数f(x)是减函数
所以函数f(x)的单调递减区间是[3π/8 +kπ，7π/8 +kπ]，k属于Z

Answer 2

f(x)=2sin²x+2sinxcosx
=1-cos2x+sin2x
=1+√2*sin(2x-π/4)
∴f(x)的最小值周期T=2π/2=π
当π/2+2kπ≤2x-π/4≤3π/2+2kπ即3π/8 +kπ≤2x-π/4≤7π/8+kπ(K∈Z)时，函数f(x)是减函数
∴函数f(x)的单调递减区间是[3π/8+kπ,7π/8+kπ](K∈Z)