Question

已知△abc的内角a,b,c的对边分别为a,b,c,-2sin²c+cosC+1=0 ,且c=3

已知△abc的内角a,b,c的对边分别为a,b,c,-2sin²C+cosC+1=0 ,且c=3
（1）求∠C
（2）若sinB-2sinA=0,求a,b的值

Answer 1

(1)
-2(sinC)^2+cosC+1 =0
-2(1-(cosC)^2)+cosC+1 =0
2(cosC)^2+cosC-1 =0
(2cosC-1)(cosC +1) =0
cosC =1/2
C= π/3
(2)
sinB-2sinA=0
sin(2π/3-A)-2sinA=0
(√3/2)cosA+(1/2)sinA -2sinA=0
(√3/2)cosA= (3/2)sinA
tanA = √3/3
A =π/6
=> B=π/2

c=3
c/sinC = a/sinA
3/(√3/2) = a/(1/2)
a=√3

c/sinC = b/sinB
3/(√3/2) = b
b= 2√3