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