huayanxi
幼苗
共回答了14个问题采纳率:85.7% 举报
y=-cosx^2+cosx+7/4
=-(cosx^2-cosx)+7/4
=-(cosx^2-cosx+1/4)+7/4+1/4
=-(cosx-1/2)^2+2
当cosx=1/2时,y取最大值,ymax=2
当cosx=-1时,y取最小值,ymin=-1/4
f(x)=sin^2wx+√3*sinwx*sin[(π/2)+wx]
=(1-cos2wx)/2+√3*sinwx*coswx
=1/2+√3/2*sin2wx
=√3/2*sin2wx-1/2*cos2wx+1/2
=sin2wxcosπ/6-cos2wxsinπ/6+1/2
=sin(2wx-π/6)+1/2
T=2π/2w=π
w=1
f(x)=sin(2x-π/6)+1/2
x∈[0,2π/3]
2x∈[0,4π/3]
2x-π/6∈[-π/6,7π/6]
sin(2x-π/6)∈[-1/2,1]
sin(2x-π/6)+1/2∈[0.3/2]
即f(x)∈[0.3/2]
f(x)的增区间:
2kπ-π/2
1年前
3