Cos4x+10tgx/(1+tg²x)=3;⇒
10tgx/(1+tg²x)=5·2tgx/(1+tg²x)=5sin2x;⇒
cos²2x-sin²2x+5sin2x=3(cos²2x+sin²2x);⇒
2cos²2x+4sin²2x-5sin2x=0;⇒
2+2sin²2x-5sin2x=0;⇒
sin2x=t;⇒-1≤t≤1;
2t²-5t+2=0;
t₁,₂=[5⁺₋√(25-16)]/4=(5⁺₋3)/4;
t₁=8/4=2>1;-корней нет;
t₂=-1/2;⇒sin2x=-1/2;⇒2x=-π/6+2kπ;k∈Z;⇒x=-π/12+kπ;k∈Z;
2x=7π/6+2kπ;k∈Z;x=7π/12+kπ;k∈Z;
x[-π/2;π/2;]⇒ответ:x=-π/12+kπ;