=sin(2x+π/6)+cos2x
=√3/2*sin2x+1/2*cos2x+cos2x
= √3/2*sin2x+3/2*cos2x
=√3*(1/2*sin2x+√3/2*cos2x)
=√3sin(2x+π/3)
單調(diào)遞增區(qū)域?yàn)椋?br/>-π/2+2kπ≤2x+π/3≤π/2+2kπ,k為整數(shù)
-5π/6+2kπ≤2x≤π/6+2kπ,k為整數(shù)
-5π/12+kπ≤x≤π/12+kπ,k為整數(shù)
則函數(shù)f(x)的單調(diào)增區(qū)間為：[-5π/12+kπ,π/12+kπ],k為整數(shù)(2)在三角形abc中 內(nèi)角abc的對(duì)邊分別為abc 已知fx=跟號(hào)3/2,a=2,sinb=3/5,求三角形abc的面積在△ABC中∵f(A)=√3sin(2A+π/3)=√3/2∴sin(2A+π/3)=1/2∴2A+π/3=π/6+2kπ或5π/6+2kπ，k為整數(shù)即A=-π/12+kπ或π/4+kπ，k為整數(shù)又∵sinB=3/5，1/2<3/5<√2/2∴π/6