2sin^2+cos4x=0 2cos^2x-sinx-2=0

Question

2sin^2+cos4x=0

2cos^2x-sinx-2=0

Answer 1

1. 2sin²x+cos4x=0

+ 2cos²2x -1=0

1-cos2x+2cos²2x-1=0

2cos²2x - cos2x=0

cos2x(2cos2x-1)=0

cos 2x₁=0 ⇒ 2x=π/2+πk ⇒ x=π/4+πk/2, где к∈Ζ

cos2x₂=1/2 ⇒2x=±π/3+2πn ⇒x=±π/6+πn, к∈Ζ

ответ - π/4+πk/2, где к∈Ζ и ±π/6+πn, к∈Ζ