Question

已知函数f(x)=根号3sin2x+2cos^2x+1

(1)求函数f(x)的单调递增区间
(2)设三角形ABC的内角A,B,C对边分别为a,b,c 且c=根号3,f(C)=3,若2sinA=sinB,求a,b的值

Answer 1

1).f(x)=(√3)sin2x+cos2x+2=2sin(2x+π/6)+2
单增枣拦液区间[-π/衡好3+kπ，π/6+kπ]
(2)f(C)=2sin(2C+π/6)+2=3，凳物故C=π/3
a/b=sinA/sinB=1/2 (1)
a²+b²-ab=3 (2)
(1)(2)联立 解得a=1，b=2，c=√3.