Question

$(2sin^2x) sqrt[4]{cosx} = sqrt[4]{cosx}$

Answer 1

ОДЗ
cosx≥0⇒x∈[-π/2+2πn;π/2+2πn
[2sin²x=0⇒sinx=0⇒x=πn,n∈z+ОДЗ⇒x=2πn,n∈z
[cox=0⇒x=π/2+2πn,n∈z