sqrt表示根号下；首先将原函数化成正弦型函数f(x)=sin2x+cos2x+2=(sqrt2)*(sin2x*(sqrt2)/2 + cos2x*(sqrt2)/2 )+2=(sqrt2)*sin(2x+ π/4）+2,当ωx+φ属于[ 2kπ-π/2 ,2kπ+π/2 ],函数单调递增,也即2x+ π/4属于[ 2kπ-π/2 ,2kπ+π/2 ],也即2x属于2kπ- 3π/4,2kπ+ π/4单调递增.所以x属于[ kπ-3π/8,kπ+π/8 ] 如果化正弦型函数有疑问的话可以追问.希望对你有所帮助!