题目

函数y=sin^2x+2sinxcosx+3cos^2x 1求函数最小正周期 2求函数的单调递增区间 3当x取什么值 函数取最大值

提问时间：2020-11-05

答案

y=sin^2x+2sinxcosx+3cos^2x =sin^2x+3cos^2x+2sinxcosx =2cos²+1+sin2x =cos2x+2+sin2x =√2sin(2x+π/4)+2最小正周期为 2π/2=π单调增区间2x+π/4∈[2kπ-π/2,2kπ+π/2]x∈[kπ-3π/8,kπ+π/8]单调减区...