Question

在三角形ABC中,内角ABC的对边分别为abc,且c/(a+b) + b/(a+c)=1,

1.求A
2.设f(x)=sin(A-2x)+2sinxcosx,x属于[0,兀],若f(x)大于等于1/2,求x的取值范围
要详细过程,谢谢

Answer 1

答：

1）
三角形ABC满足：c/(a+b)+b/(a+c)=1
变形得：
ac+c^2+ab+b^2=a^2+ac+ab+bc
a^2=b^2+c^2-bc
=b^2+c^2-2bccosA
所以：
cosA=1/2
A=60°
2）
f(x)=sin(A-2x)+2sinxcosx
=sin(π/3-2x)+sin2x
=2sin(π/6)cos(2x-π/6)
=cos(2x-π/6)>=1/2
0<=x<=π，0<=2x<=2π
-π/6<=2x-π/6<=11π/6
所以：
-π/6<=2x-π/6<=π/3或者5π/3<=2x-π/6<=11π/6
解得：0<=x<=π/4或者11π/12<=x<=π