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