Решите уравнение: l cosx l = cosx-2sinx
Cosx=cosx-2sinx sinx=0 cosx>=0 x=2Пk -cosx=cos-2sinx cosx=sinx cosx<0<br>x=5П/4+2Пk
Сosx<0⇒x∈(π/2+2πn,3π/2+2πn)<br>-cosx=cosx-2sinx 2sinx-2cosx=0/cosx 2tgx-2=0 tgx=1 x=π/4+πn +x∈(π/2+2πn,3π/2+2πn) х=5π/4+2πn,n∈z 2)cosx≥0⇒x∈[-π/2+2πn;π/2+2πn,n∈z] cosx=cosx-2sinx sinx=0 x=πn +x∈[-π/2+2πn;π/2+2πn,n∈z] x=2πn,n∈z