f(x)=根号3sinxcosx+cosx^2,f(x)=1-根号3/2,求x

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/07 05:14:05
f(x)=根号3sinxcosx+cosx^2,f(x)=1-根号3/2,求x

f(x)=根号3sinxcosx+cosx^2,f(x)=1-根号3/2,求x
f(x)=根号3sinxcosx+cosx^2,f(x)=1-根号3/2,求x

f(x)=根号3sinxcosx+cosx^2,f(x)=1-根号3/2,求x
f(x)=√3sinxcosx+cosx^2
=√3/2sin2x+(1+cos2x)/2
=√3/2sin2x+cos2x/2+1/2
=sin2xcosπ/6+cos2xsinπ/6+1/2
=sin(2x+π/6)+1/2
sin(2x+π/6)+1/2=(1-√3)/2
sin(2x+π/6)=-√3/2
2x+π/6=2kπ-π/3或2x+π/6=2kπ+4π/3
2x=2kπ-π/2或2x=2kπ+7π/6
x=kπ-π/4或x=kπ+7π/12

上面的会打的不错哦,我们信仰信春哥