已知向量a(cosa,sina),b(cosx,sinx),c=(sinx+2sina,cosx+2cosa),其中0

1)
(a=π/4
c=(sinx+√2,cosx+√2)
f(x)=b●c=cosx(sinx+√2)+sinx(cosx+√2)
=2sinxcosx+√2(sinx+cosx)
设 sinx+cosx=t
∴t²=(sinx+cosx)²=1+2sinxcosx
∴ 2sinxcosx=t²-1
又t=sinx+cosx=√2sin(x+π/4)∈[-√2,√2]
∴f(x)=g(t)=t²-1+√2t=(t+√2/2)²-3/2
∴t=-√2/2时,g（t),即f(x) 取得最小值 -3/2
此时,√2sin(x+π/4）=-√2/2
∴sin(x+π/4）=-1/2
∵0