(1)â³ABCä¸,(2c-(â3)a)cosB-(â3)bcosA=0
ç±æ£å¼¦å®çå¾ (2sinC-(â3)sinA)cosB-(â3)sinBcosA=0
2sinCcosB=(â3)(sinAcosB+sinBcosA)=(â3)sin(A+B)=(â3)sin(Ï-C)=(â3)sinC
å³2sinCcosB=(â3)sinC èsinC>0
cosB=(â3)/2,Bæ¯â³ABCå
è§,å¾B=Ï/6
A=Ï-2*(Ï/6)=2Ï/3
â³ABCçé¢ç§¯S=(1/2)(â3)^2*sin(2Ï/3)=(3â3)/4
(2)ç±(1)å¯å¾BC=3,Aç¹çºµåæ æ¯(â3)/2
f(x)=λsin(Ïx+Ï/3) (λ>0,Ï>0)
T/2=3,å¨æT=6, Ï=2Ï/T=Ï/3,λ=(â3)/2
f(x)=(â3/2)sin((Ï/3)x+Ï/3)
ç±kÏ-Ï/2â¤(Ï/3)x+Ï/3â¤kÏ+Ï/2,kâZ
3k-5/2â¤xâ¤3k+1/2,kâZ
æ以f(x)çåå¢åºé´æ¯[3k-5/2,3k+1/2],kâZ
å¸æè½å¸®å°ä½ ï¼
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