2cos^2x+根号3sin2x+m=2sin（2x+π/6）+m+1,怎么做的?

题目

2cos^2x+根号3sin2x+m=2sin（2x+π/6）+m+1,怎么做的?
2cos^2x+根号3sin2x+m=2sin（2x+π/6）+m+1,左边是怎么算到右边的,降幂公式算完之后我得到的是cos2x+根号3sin2x+m+1,之后呢

答案

cos2x+根号3sin2x+m+1
=2(1/2*cos2x+√3/2*sin2x)+m+1
=2(sinπ/6cos2x+sin2xcosπ/6)+m+1
=2sin（2x+π/6）+m+1