Ответы и объяснения

  • Участник Знаний
2014-03-31T00:13:26+00:00
1)cosx≥0⇒x∈[-π/2+2πn;π/2+2πn]
cosx=cosx-2sinx
2sinx=0⇒sinx=0⇒x=πn U x∈[-π/2+2πn;π/2+2πn]⇒x=2πn
2)cosx<0⇒x∈(π/2+2πn;3π/2+2πn)
-cosx=cosx-2sinx
2sinx-2cosx=0
sinx-sin(π/2-x)=0
2sin(x-π/4)cosπ/4=0
sin(x-π/4)=πn
x=π/4+πn U x∈(π/2+2πn;3π/2+2πn)⇒5π/4+2πn