Question

решить тригонометрическое уравнение
cos^2*2x=cos^2*4x

Answer 1

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.