f(x)=2sin(wx-π/3)coswx+2cos(2wx+π/6),其中w>0,若fx满足f(π+x)=f(π-x)w∈(0.5,1),求函数fx的单调递减区间
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f(x)=2sin(wx-π/3)coswx+2cos(2wx+π/6),其中w>0,若fx满足f(π+x)=f(π-x)w∈(0.5,1),求函数fx的单调递减区间f(x)=2sin(wx-π/3)
f(x)=2sin(wx-π/3)coswx+2cos(2wx+π/6),其中w>0,若fx满足f(π+x)=f(π-x)w∈(0.5,1),求函数fx的单调递减区间
f(x)=2sin(wx-π/3)coswx+2cos(2wx+π/6),其中w>0,
若fx满足f(π+x)=f(π-x)w∈(0.5,1),求函数fx的单调递减区间
f(x)=2sin(wx-π/3)coswx+2cos(2wx+π/6),其中w>0,若fx满足f(π+x)=f(π-x)w∈(0.5,1),求函数fx的单调递减区间
f(x)=2sin(wx-π/3)coswx+2cos(2wx+π/6)
=2(sinwxcosπ/3-coswxsinπ/3)coswx+2cos(2wx+π/6)
=sinwxcoswx-√3(coswx)^2+2cos(2wx+π/6)√3cos2wx-sin2wx
=1/2sin2wx-√3/2(1+cos2wx)+2cos(2wx+π/6)√3cos2wx-sin2wx
=-(√3/2cos2wx-1/2sin2wx)-√3/2+2cos(2wx+π/6)
=-cos(2wx+π/6)-√3/2+2cos(2wx+π/6)
=cos(2wx+π/6)-√3/2
你把楼上的括号里的放进COS的递减区间
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