Question

8cos^4x-8cos^2x-cosx+1=0

Answer 1

8cos²x(cos²x-1)-(cosx-1)=0
8cos²x(cosx-1)(cosx+1)-(cosx-1)=0
(cosx-1)(8cos³x+8cos²x-1)=0
cosx-1=0⇒cosx=1⇒x=2πn,n∈z
8cos³x+8cos²x-1=0 дейст.корней нет