题目

函数y=3sin^2x+cos2x的最小正周期为?
y=3sin^2x+cos2x
=3-3(cos2x+1)/2+cos2x这步是怎么做的,
=-cos2x/2+3/2
函数y=3sin^2x+cos2x的最小正周期为?重新做一次，写上全部步骤

提问时间：2020-11-08

答案

1-0.5（cos2x+1）=1-0.5(cos^2 x- sin^2 x + sin^2 x+cos^2 x)=1-cos^2 x=sin^2 x3sin^2x=3-3(cos2x+1)/2 y=3sin^2x+cos2x=3(1-cos^2 x)+cos2x=3[1-0.5(cos^2 x- sin^2 x + sin^2 x+cos^2 x)]+cos2x=3-3(cos2x+1)/2+...