Question

$tg^2(x)-3tg(x)+ frac{2sin(x)}{cos^2(x)}=frac{3}{cos^3(x)} -frac{1}{cos^4(x)}$

Answer 1

$tg^2(x)-3tg(x)+ frac{2sin(x)}{cos^3(x)} =frac{3}{cos^2(x)} -frac{1}{cos^4(x)} \x neq frac{pi}{2} +k pi\tg^2(x)+1=frac{1}{cos^2(x)} \(tg^2(x)+1)^2=frac{1}{cos^4(x)} \tg^2(x)-3tg(x)+2tg(x)*frac{1}{cos^2(x)} =frac{3}{cos^2(x)}-frac{1}{cos^4(x)} \tg(x)=t\t^2-3t+2t(t^2+1)=3(t^2+1)-(t^2+1)^2\t^2-3t+2t^3+2t=3t^2+3-t^4-2t^2-1\t^4+2t^3-t-2=0\t(t^3-1)+2(t^3-1)=0\(t+2)(t^3-1)=0\t_1=1\t_2=-2\tg(x)=1\x=frac{pi}{4}+pi k\tg(x)=-2\x=-arctg(2)+pi k$
k∈Z