f(x)=sin^4x+2sinxcosx+cos^4x的最小值

f(x)=sin^4x+2sinxcosx+cos^4x
=(sin^2x+cos^2x)^2+sin2x-1/2*sin^2(2x)
=1+sin2x-1/2*sin^2(2x)
=3/2-1/2(1-sin2x)^2
-1≤sin2x≤1,当sim2x=-1时
f(x)有最小值-1/2