Question

Помогите пожалуйста
1.) 3tgx=ctgx
2.) tg в квадрате x - 5 = 4 tg x

Answer 1

$1) 3tgx=ctgx \ frac{3sinx}{cosx}- frac{cosx}{sinx}=0 \ 3sin^{2}x-cos^{2}x=0 \ 3(1-cos^{2}x)-cos^{2}x=0 \ 3-3cos^{2}x-cos^{2}x=0 \ -4cos^{2}x+3=0 \ -4cos^{2}x=-3 \ cos^{2}x=3/4 \ cosx= frac{ sqrt{3}}{2} \ x=arccosfrac{sqrt{3}}{2}+2 pi n \ x= frac{ pi }{6}+2 pi n$$2) \ tg^{2}x-5=4tgx \ tg^{2}x-4tgx-5=0 \ tgx=t \ t^{2}-4t-5=0 \ D=(-4)^{2}-4*(-5)=16+20=36 \ t_{1}=4+6/2=5 \ t_{2}=4-6/2=-1 \ tgx=5 \ x=arctg5+ pi n \ tgx=(-1) \ x=- frac{ pi }{4} + pi n$