已知复数z1=cos2x+λi,z2=m+(sin2x-
3
(Ⅰ)若λ=0时,且
(Ⅱ)设λ=f(x),求f(x)的单调递增区间.
李志华回答:
(Ⅰ)复数z1=cos2x+λi,z2=m+(sin2x-3m)i(λ,m,x∈R),且z1=z2,cos2x=mλ=sin2x-3cos2x=2sin(2x-π3),所以sin(2x-π3)=0,2x-π3=kπ,x=kπ2+π6,k∈Z,又π2<x<π,所以x=2π3;(Ⅱ)由2kπ−π...
相关文章:
已知复数z1=cos2x+λi,z2=m+(sin2x-
3
(Ⅰ)若λ=0时,且
(Ⅱ)设λ=f(x),求f(x)的单调递增区间.
李志华回答:
(Ⅰ)复数z1=cos2x+λi,z2=m+(sin2x-3m)i(λ,m,x∈R),且z1=z2,cos2x=mλ=sin2x-3cos2x=2sin(2x-π3),所以sin(2x-π3)=0,2x-π3=kπ,x=kπ2+π6,k∈Z,又π2<x<π,所以x=2π3;(Ⅱ)由2kπ−π...