这道三角函数题怎么解

2024-12-20 08:02:08
推荐回答(3个)
回答1:

f(x)=sin2x-根号下3cos2x+1
=2(sin2x×1/2-cos2x×根号下3/2)+1
=2(sin2xcosπ/3-cos2xsinπ/3)+1
=2sin(2x-π/3)+1

回答2:

f(x)=sin2x-√3cos2x+1
f(x)=2(1/2sin2x-√3/2cos2x)+1
f(x)=2(cosπ/3sin2x-sinπ/3cos2x)+1
f(x)=2cos(2s-π/3)+1

回答3:

不知