已知函数 f ( x )=sin +2cos 2 x -1( x ∈R).

1个回答

  • (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