向量m=(2sinx,1),n=(√3cosx,2cos2x),函数f(x)=mn-t.(1)若方程f(x)=0在0

向量m=(2sinx,1),n=(√3cosx,2cos2x),函数f(x)=mn-t.(1)若方程f(x)=0在0
向量m=(2sinx,1),n=(√3cosx,2cos2x),函数f(x)=mn-t.(1)若方程f(x)=0在0<=x<=π/2上有解,求t取值范围；
(2)在三角形ABC中,abc分别是ABC所对的边,当(1)中的t取最大值且f(A)=-1,b+c=2时,求a的最小值.

向量m=(2sinx,1),n=(√3cosx,2cos2x),函数f(x)=mn-t.(1)若方程f(x)=0在0
1:f(x)=2√3sinxcosx+2cos2x-t=√3sin2x+2con2x-t=2sin(2x+π/6)-t
f(x)=0,2sin(2x+π/6)＝t,
0

f(x)=2√3sinxcosx+2cos2x-t=√3sin2x+2con2x+1-t=2sin(2x+π/6)+1-t
f(x)=0，2sin(2x+π/6)+1＝t,
0<=x<=π/2, π/6<=2x+π/6<=7π/6,
0<=2sin(2x+π/6)+1<=3,
0<=t<=3
2. f(A)=-1,故sin(2A+π/6)＝1/2,得A=0（...

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f(x)=2√3sinxcosx+2cos2x-t=√3sin2x+2con2x+1-t=2sin(2x+π/6)+1-t
f(x)=0，2sin(2x+π/6)+1＝t,
0<=x<=π/2, π/6<=2x+π/6<=7π/6,
0<=2sin(2x+π/6)+1<=3,
0<=t<=3
2. f(A)=-1,故sin(2A+π/6)＝1/2,得A=0（舍去）或A=π/3
a^2=b^2+c^2-2bccosA
=(b+c)^2-2bc-2bc*(1/2)
=4-3bc
b+c=2≧√bc(基本不等式) 所以bc≤1 4-3bc≤1 a≤1 所以a(max)=1

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