(1)
( k ∈Z)(2) a =3
f ( x )=sin
+2cos 2x -1=-
cos 2 x +
sin 2 x +cos 2 x =
cos 2 x +
sin 2 x =sin
.
(1)最小正周期 T =
=π,由2 k π-
≤2 x +
≤2 k π+
( k ∈Z),得 k π-
≤ x ≤ k π+
( k ∈Z),所以 f ( x )的单调递增区间为
( k ∈Z).
(2)由 f ( A )=sin
=
得2 A +
=
+2 k π或
+2 k π( k ∈Z),即 A = k π或 A =
+ k π,又 A 为△ ABC 的内角,所以 A =
.
又因为 b , a , c 成等差数列,所以2 a = b + c .
∵
·
= bc cos A =
bc =9,∴ bc =18,∴cos A =
=
-1=
-1=
-1.∴ a =3