急！！！已知函数f(x)=sin(2x+π⼀3)......

f(x)=sin(2x+π/3)-√3*sin²x+sinxcosx+(√3)/2
= sin(2x+π/3)-√3*sin²x+1/2*sin2x+(√3)/2
= 1/2*sin2x+ (√3)/2*cos2x-√3*sin²x+1/2*sin2x+(√3)/2
= sin2x+ (√3)/2*cos2x-√3*sin²x+(√3)/2
= sin2x+ (√3)/2*(1-2 sin²x )-√3*sin²x+(√3)/2
= sin2x+ (√3)/2*-√3sin²x -√3*sin²x+(√3)/2
= sin2x-2√3* sin²x+√3
= sin2x-2√3*(1- cos2x)/2+√3
= sin2x+√3* cos2x-√3
=1/2* sin(2x+π/3)-√3.
供参考，你再找一下更简单的方法吧。

原式=sin(2x+π/3) + (sin2x)/2 + √3/2*[1-2(sinx)^2]
=sin(2x+π/3) + (sin2x)/2 + √3/2*cos2x
=sin(2x+π/3) + sin(2x+π/3)
=2sin(2x+π/3)