f(x)=2√3sinxcosx+2cos²x-t (1)若方程f(x)=0在x∈[0,π/2]上有解,求t的范围

Question

（2）在△ABC中,a,b,c分别是A,B,C所对的边,若t=3,且f(A)=-1,b+c=2,求a的最小值

Answer 1

解：
(1)
f(x)=2√3sinxcosx+2cos²x-t
=√3sin(2x)+1+cos(2x)-t
=√3sin(2x)+cos(2x)+1-t
=2[(√3/2)sin(2x)+(1/2)cos(2x)]+1-t
=2sin(2x+π/6)+1-t
f(x)=0
2sin(2x+π/6)+1-t=0
sin(2x+π/6)=(t-1)/2
x∈[0，π/2]，则π/6≤2x+π/6≤7π/6
-½≤sin(2x+π/6)≤1
要方程有解，-½≤(t-1)/2≤1
0≤t≤3
t的取值范围为[0，3]
(2)
t=3，f(x)=2sin(2x+π/6)+1-t=2sin(2x+π/6)+1-3=2sin(2x+π/6)-2
f(A)=-1
2sin(2A+π/6)-2=-1
sin(2A+π/6)=½
A为三角形内角，0<A<π
π/6<2A+π/6<13π/6
2A+π/6=5π/6
A=π/3
由余弦定理得：
a²=b²+c²-2bccosA
=(b+c)²-2bc-2bc·cos(π/3)
=2²-2bc-2bc·½
=4-3bc
b+c=2
由均值不等式得：(b+c)²≥4bc
bc≤(b+c)²/4=2²/4=1
a²≥4-3=1
a≥1
a的最小值为1