1,f(x)=sin(πx/4-π/6)-2cos^2πx/8+1
=[(√3/2)sinπx/4-(1/2)cosπx/4]-（1+cosπx/4）+1
=(√3/2)sinπx/4-(3/2)cosπx/4
=√3[(1/2)sinπx/4-(√3/2)cosπx/4]
=√3sin(πx/4-π/3),
所以T=2π÷π/4=8.
2,sinx的单调递减区间为[2kπ+π/2,2kπ+3π/2]
πx/4-π/3∈[2kπ+π/2,2kπ+3π/2]得x∈[8k+10/3,8k+22/3]