已知函数y=sin2x+2sinxcosx+3cos2xx属于R.（1）当x为何值时,函数有最大值?...

y=sin(2x)+2sinxcosx+3cos(2x)
=sin(2x)+2sin(2x)+3cos(2x)
=3sin(2x)+3cos(2x)
=3√2sin(2x+π/4)
当2x+π/4=2kπ+π/2时,即当x=(8k+1)π/8时,函数有最大值ymax=3√2
2kπ-π/2≤2x+π/4≤2kπ+π/2时,函数单调递增,此时(8k-3)π/2≤x≤(8k+1)π/8
2kπ+π/2≤2x+π/4≤2kπ+3π/2时,函数单调递增,此时(8k+1)π/2≤x≤(8k+7)π/8
函数的单调递增区间为[(8k-3)π/2,(8k+1)π/8],单调递减区间为[(8k+1)π/2,(8k+7)π/8].