解法如下,
f(A)=2sin(2A+π/6)=1
sin(2A+π/6)=1/2
2A+π/6=π/6或2A+π/6=5π/6
得:A=0【舍去】或者A=π/3
又:a=√3、A=π/3,则:
a²=b²+c²-2bccosA=b²+c²-ba
b²+c²-bc=3
bc=(1/3)[(b+c)²-3]
因为:b+c≥2√(bc),即:
bc≤(1/4)(b+c)
则:
(1/4)(b+c)²≥(1/3)(b+c)²-1
(b+c)²≤12
b+c≤2√3
即:b+c最大值是2√3
所以
v3