Question

$tg^{2}x+cos6x*tg^{2}x-cos6x=1$

Answer 1

$tg^2x+cos6x*tg^2x-cos 6x=1\ tg^2x(1+cos 6x)-(1+cos 6x)=0\ (1+cos 6x)(tg^2x-1)=0\ 1+cos 6x=0\ x= frac{pi}{6} + frac{pi n}{3},n in Z\ \ tg^2x-1=0\ tgx=pm1\ x=pm frac{pi}{4}+ pi k,k in Z$