Question

Решите уравнение:
l cosx l = cosx-2sinx

Answer 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

Answer 2

cosx=cosx-2sinx
sinx=0  cosx>=0
x=2Пk

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