y=1+sin2x+2cos^2x-1+1=sin2x+cos2x+2=√2sin(2x+π/4)+22x+π/4=π/2+2kπ（k∈Z）2x=π/4+2kπx=π/8+kπ2x+π/4=3π/2+2kπ（k∈Z）2x=5π/4+2kπx=5π/8+kπ递减区间[π/8+kπ,5π/8+kπ]（k∈Z）y（max）=√2+2...