Решить тригонометрическое уравнение

cos^2*2x=cos^2*4x

Cos²2x=cos²4x; ⇔cos²2x = (2cos²2x-1) ²⇔cos²2x=4cos⁴2x-4cos²2x+1; ⇔

4cos⁴2x-5cos²2x+1=0; ⇒cos²2x=t; cos²2x>0; cos²2x≤1; ⇒t²-3

4t²-5t+1=0;

t₁,₂ = (5⁺₋√25-4·4·1) / 8 = (5⁺₋3) / 8;

t₁ = (5+3) / 8=1; ⇒cos²2x=1;

cos2x=+1; 2x=2kπ; k∈Z; x=kπ; k∈Z;

cos2x=-1; 2x=π+2kπ; k∈Z; x=π/2+kπ; k∈Z;

t₂=1/4; ⇒cos²2x=1/4; cos2x=⁺₋1/2;

cos2x=1/2; ⇒2x=⁺₋π/3+2kπ; k∈Z; x=⁺₋π/6+kπ; k∈Z;

cos2x=-1/2; 2x=⁺₋2π/3+2kπ; k∈Z; x=⁺₋π/3+kπ; k∈Z.