设函数f(x)=cos(x+2π/3)+2cos²x/2,x属于R
问题描述：
设函数f(x)=cos(x+2π/3)+2cos²x/2,x属于R
.记△ABC内角A B C的对边长分别为a,b,c,若f(B)=1,b=1,c=根号三,求a
答
f(B)=cos(B+2π/3)+2cos²B/2=1推出cos(B+2π/3)= -cosB所以B+2π/3+B=π推出B=π/6
cosB=(a²+c²-b²)/2*a*c=（a²+3-1)/2*a*根号3=根号3/2推出a=1或2~