f(x)=sin2x -2cos²x+3
=sin2x-[1+cos(2x)]+3
=sin2x-cos(2x)+2
=√2sin(2x- π/4)+2
sin(2x-π/4)=1时,f(x)有最大值[f(x)]max=2+√2
此时,2x-π/4=2kπ+π/2 (k∈Z)
x=kπ+3π/8 (k∈Z)
x的集合为{x|x=kπ+3π/8 ,k∈Z}
2kπ-π/2≤2x- π/4≤2kπ+π/2 (k∈Z)时,函数单调递增.
kπ-π/8≤x≤kπ+3π/8 (k∈Z)
函数的单调递增区间为[kπ-π/8,kπ+3π/8] (k∈Z)
f(x)>3
√2sin(2x- π/4)+2>3
sin(2x-π/4)>√2/2
2kπ+3π/4