Решите пожалуйста
cos^-2 (2x) -sin^-2 (2x)= 8/3

1/cos²2x-1/sin²2x=8/3;⇒(sin²2x-cos²2x)/(cos²2x·sin²2x)=8/3;⇒-cos4x=8/3·(cos²2x·sin²2x);  -cos4x=8/3·1/4·sin²4x;⇒cos4x+2/3sin²4x=0;⇒cos4x+2/3(1-cos²4x)=0;⇒2/3cos²4x-cos4x-2/3=0;⇒cos4x=t;-1<t<1;2/3t²-t-2/3=0;t₁,₂=(1⁺₋√(1+4·2/3·2/3))/(4/3)=(1⁺₋5/3)/(4/3)t₁=(1+5/3)/(4/3)=2⇒корней нет,т.к 2>1t₂=(1-5/3)/(4/3)=-1/2;⇒cos4x=-1/2;⇒ 4x=⁺₋2π/3+2kπ;k∈Z;x=⁺₋π/6+kπ/2;k∈Z;