(1)若f(1)=0,且B-C=,求角C;
(2)若f(2)=0,求角C的取值范围.
(1)若f(1)=0,且B-C=,求角C;
(2)若f(2)=0,求角C的取值范围.
解:(1)由f(1)=0,得a2-a2+b2-4c2=0,∴b=2c.
又由正弦定理,得b=2RsinB,c=2RsinC,将其代入上式,得sinB=2sinC.
∵B-C=,∴B=
+C,将其代入上式,得sin(
+C)=2sinC.
∴sincosC+cos
sinC=2sinC,整理得,
sinC=cosC.
∴tanC=.
∵角C是三角形的内角,∴C=.
(2)∵f(2)=0,∴4a2-2a2+2b2-4c2=0,即a2+b2-2c2=0.
由余弦定理,得cosC=
==
(当且仅当a=b时取等号).
∴cosC≥,∠C是锐角.
又∵余弦函数在(0,)上递减,∴0<C≤
.