已知函数f(x)=2根号3sinxcosx+2cos^2(x)-t在三角形abc中f(A)=-1,b+c=2,若t=3求a的最小值
解析:∵函数f(x)=2√3sinxcosx+2cos^2(x)-t=√3sin2x+cos2x+1-t
=2sin(2x+π/6)+1-t
∵f(A)=-1==> f(A)=2sin(2A+π/6)+1-t=-1==>sin(2A+π/6)=(t-2)/2
当t=3时,sin(2A+π/6)=1/2==>2A+π/6=5π/6==>A=π/3
∵b+c=2==>b=2-c
由余弦定理a^2=c^2+(2-c)^2-2c(c-2)*1/2=c^2-2c+4=(c-1)^2+3
∴当b=c=1时,a取最小值√3