y=cos^2 x-cos x+2=(cosx-1/2)²+7/4设t=cosx∈[-1,1]y=(t-1/2)²+7/4当t∈[-1,-1/2]时,y=(t-1/2)²+7/4递减此时x∈[2π/3+2kπ,π+2kπ],k∈Z,t=cosx递减∴[2π/3+2kπ,π+2kπ],为原函数的递增区间x∈...对不起,你的答案好像错了,我套进去不成立呀确实有点疏漏当t∈[-1,1/2]时， y=(t-1/2)²+7/4递减 此时x∈[π/3+2kπ,π+2kπ],k∈Z,t=cosx递减∴[π/3+2kπ,π+2kπ],为原函数的递增区间 x∈[π+2kπ,5π/3+2kπ],k∈Z,t=cosx递增∴[π+2kπ,5π/3+2kπ],为原函数的递减区间当t∈[1/2,1]时， y=(t-1/2)²+7/4递增此时x∈[5π/3+2kπ,2π+2kπ],k∈Z,t=cosx递增∴[5π/3+2kπ,2π+2kπ],k∈Z 为原函数的递增区间 x∈[2kπ, π/3+2kπ],k∈Z,t=cosx递减∴[2kπ, π/3+2kπ],k∈Z为原函数的递减区间∴原函数的递减区间是[2kπ, π/3+2kπ],[π+2kπ,5π/3+2kπ],k∈Z 原函数的递增区间是[5π/3+2kπ,2π+2kπ], [π/3+2kπ,π+2kπ], k∈Z