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⼀2上有解，求t取值范围；

1: f(x)=2√3sinxcosx+2cos2x-t=√3sin2x+2con2x-t=2sin(2x+π/6)-t
f(x)=0，2sin(2x+π/6)＝t,
0<=x<=π/2, π/6<=2x+π/6<=7π/6,
-2<=2sin(2x+π/6)<=2,
-2<=t<=2
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
bc<=(b+c)^2/4,当b=c时等号成立
bc(max)=1,a^2(min)=1,a(min)=1

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