求这个函数的周期,谢谢!

y=3(sinx)^2+sinxcosx+4(cosx)^2
=3(sinx)^2+sinxcosx+3(cosx)^2+(cosx)^2
=3+sinxcosx+(cosx)^2
=3+(1/2)sin2x+(cos2x+1)/2
=(1/2)(sin2x+cos2x)+7/2
=√2/2*sin(2x+π/4)+7/2
T=2π/2=π

y=3(sinx)^2+sinxcosx+4(cosx)^2

=3+sinxcosx+(cosx)^2

=3+(1/2)sin(2x)+(1/2)cos(2x)+1/2

=7/2+〔(根号2）/2〕sin(2x+兀/4)

所以函数Y的周期为2兀／2＝兀

3(sinx)^2+sinxcosx+4(cosx)^2=3[(sinx)^2+(cosx)^2]+0.5*sin2x+(sinx)^2=3+0.5sin2x+(cos2x-1)/2

所以很明显,周期t=2pi/2=pi

3(sinx)^2+sinxcosx+4(cosx)^2=
3(sinx)^2+3(cosx)^2+(cosx)^2+sinxcosx=
1+(cosx)^2+sinxcosx=1+cos(2x)-1+sinxcosx
=cos(2x)+sin(2x)
又有两个的周期都为π所以周期为π

解：
y=3(sinx)^2+sinxcosx+4(cosx)^2
=3/2(1-cos2x)+1/2sin2x+2(1+cox2x)
=3/2-3/2cos2x+1/2sin2x+2+2cos2x
=1/2sin2x+1/2cos2x+7/2
=√2/2sin(2x+派/4)+7/2

函数周期T=2派/2=派