Question

高分急急急！！！求解高中数学题

已知f(x)=2cos平方(x+π/3)+cos(2x-2π/3)求最小周期以及单调递增区间
要详细过程

Answer 1

f(x)=1+2cos(2x+2π/3)+cos(2x-2π/3)
=1+2cos2xcos2π/3
=1-cos2x
∵f(x)=cosx的最小正周期为2π
∴f(x)=1-cos2x的最小正周期为π
其单调递增区间即为f(x)=cos2x的单调递减区间
即[kπ,kπ+π/2]

Answer 2

f(x)=1+cos(2x+2TT/3)+cos(2x-2TT/2)=........自己化简就知道了

Answer 3

f(x)=2cos平方（x+π/3）+cos(2x-2π/3)
=cos(2x+2π/3)+1+cos(2x-2π/3)
=cos2xcos2π/3-sin2xsin2π/3+1+cos2xcos2π/3+sin2xsin2π/3
=-cos2x+1
所以周期为 π
单点增区间为[kπ，π/2+kπ]
采纳啊，楼主