Question

Помогите решить, пожалуйста! cos4x + 2cos^2x =1

Answer 1

Cos4x+2cos²x-1=0
cos4x+cos2x=0
2cos²2x-1+cos2x=0
cos2x=a
2a²+a-1=0
D=1+8=9
a1=(-1-3)/4=-1⇒cos2x=-1⇒2x=π+2πn,n∈z⇒x=π/2+πn,n∈z
a2=(-1+3)/4=1/2⇒cos2x=1/2⇒2x=+-π/3+2πk,k∈z⇒x=+-π/6+πk,k∈z