Ответ:1)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

Объяснение:

я решил вот так |cosx|=cosx-2sinx