求f(x)=sinxcosx+3(sinx)^2+2(cosx)^2的最小值

f(x)=sinxcosx+3(sinx)^2+2(cosx)^2
=sin2x/2+(sinx)^2+2
=sin2x/2-(cos2x)/2+5/2
=(√2/2)*sin[2x-π/4]+5/2
所以当x=kπ+7π/8时
f(x)的最小值=5/2-(√2/2)

f(x)=sinxcosx+3(sinx)^2+2(cosx)^2
=(√3sinx+√2cosx)^2+(1-2√6)sinxcosx
=(√3sinx+√2cosx)^2+[(1-2√6)/2]sin2x
===>f(x)min=(1-2√6)/2

两边求导令他为0得x＝3π/8
带入得最大值2^(1/2)+5/2
x＝7π/8时
带入得最小值5-2^(1/2)/2