Решите уравнение:

l cosx l = cosx-2sinx

Сosx<0⇒x∈ (π/2+2πn, 3π/2+2πn)

-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

Cosx=cosx-2sinx

sinx=0 cosx>=0

x=2 Пk

-cosx=cos-2sinx

cosx=sinx cosx<0

x=5 П/4+2 Пk