6/π?通常是π/6啊,下面我當(dāng)π/6來(lái)解吧：
f(x)=sin(2x+π/6)-sin(2x-π/6)-cos2x+1
=(sin2xcosπ/6+cos2xsinπ/6)-(sin2xcosπ/6-cos2xsinπ/6)-cos2x+1
=2cos2xsinπ/6-cos2x+1
=cos2x-cos2x+1
=1抱歉，一激動(dòng)就把題目打得穿越了。補(bǔ)充了新的題目，不嫌棄的請(qǐng)重新回答一下，有理必采納。對(duì)任意x∈R，恒有f(x)=1，是常數(shù)函數(shù)沒(méi)有最小正周期【任何正實(shí)數(shù)都是它的正周期】對(duì)任意a∈R，x=a都是它的對(duì)稱軸對(duì)任意a∈R，點(diǎn)P(a，0)都是它的對(duì)稱中心沒(méi)有嚴(yán)格單調(diào)遞增區(qū)間。抱歉題目還是打錯(cuò)了，應(yīng)該是：函數(shù)f(x)=sin(2x+π/6)+sin(2x-π/6)-cos2x+1，求f(x)的最小正周期、對(duì)稱軸、對(duì)稱中心、單調(diào)增區(qū)間。(x)=sin(2x+π/6)+sin(2x-π/6)-cos2x+1=(sin2xcosπ/6+cos2xsinπ/6)+(sin2xcosπ/6-cos2xsinπ/6)-cos2x+1=2(sin2x)cosπ/6-2cos2xsinπ/6+1=2sin(2x-π/6)+1最小正周期T=2π/2=πsin(2x-π/6)=1或-1，即2x-π/6=kπ+π/2，即x=kπ/2+π/3是對(duì)稱軸，【k是整數(shù)，下同】sin(2x-π/6)=0，即2x-π/6=kπ，即x=kπ/2+π/12，點(diǎn)P(kπ/2+π/12，1)是對(duì)稱中心sin(2x-π/6)=1，即2x-π/6=2kπ+π/2，x=kπ+π/3，T=π，單調(diào)遞增區(qū)間為[kπ-2π/3，kπ+π/3]函數(shù)的最大值3，最小值-1，即值域[-1，3]