Решите уравнение:
l cosx l = cosx-2sinx

сosx<0⇒x∈(π/2+2πn,3π/2+2πn)-cosx=cosx-2sinx2sinx-2cosx=0/cosx2tgx-2=0tgx=1x=π/4+πn +x∈(π/2+2πn,3π/2+2πn)х=5π/4+2πn,n∈z2)cosx≥0⇒x∈[-π/2+2πn;π/2+2πn,n∈z]cosx=cosx-2sinxsinx=0x=πn +x∈[-π/2+2πn;π/2+2πn,n∈z]x=2πn,n∈z

cosx=cosx-2sinxsinx=0  cosx>=0x=2Пk-cosx=cos-2sinxcosx=sinx  cosx<0x=5П/4+2Пk