Question

已知函数f(x)＝4sinωxsin^2（π/4+ωx/2）+cos2ωx（ω＞0）

已知函数f(x)＝4sinωxsin^2（π/4+ωx/2）+cos2ωx（ω＞0）1.求F(X)的最大值2.若f(x)在区间【-π/2，2π/3】上是增函数，求ω取值范围

Answer 1

解：
f(x)＝4sinωxsin^2（π/4+ωx/2）+cos2ωx
=2sinωx[1-cos(π/2+ωx)]+cos2ωx
=2sinωx[1+sinωx]+cos2ωx
=2sinωx+1
故最大值＝2＋1＝3
2、
f(x)在区间【-π/2，2π/3】上是增函数，它的最小正周期>=(2π/3+π/2)*2=7π/3
故ω<=2π/(7π//3)=6/7
ω<=6/7