y=2sin^2x-sinxcosx+3cos^2x求最小正周期,单调增区间,在[0,π/2]上的值域_数学解答

y=2sin^2x-sinxcosx+3cos^2x求最小正周期,单调增区间,在[0,π/2]上的值域

人气：192 ℃　时间：2020-01-29 23:52:00

解答

y=2(sinx)^2-sinxcosx+3(cosx)^2
=2-sin2x/2+(cosx)^2=2-sin2x/2+cos2x/2+1/2
=5/2+(√2/2)cos(2x+π/4)
π+2kπ