Question

化简fx=-根号2sin（2x+π/4）+6sinxcosx-2cos²x+1 x属于R

Answer 1

f（x）=-sin2x-cos2x+3sin2x-cos2x

=2sin2x-2cos2x

=2根号2sin(2x-π/4)

T=2π/2=π

-π/2+2kπ≤2x-π/4≤π/2+2kπ k属于Z

-π/8+kπ≤x≤3π/8+kπ

递增区间[-π/8+kπ，3π/8+kπ]

在0-π/2上最大值为2根号2，x=3π/8最小值为-2 x=0

请采纳。