Rozwiąż równanie, proszę o pomoc :  tg 4x = -1  sin 2x tgx = sinx  1+ sin 2x = cos 2x

tg 4x=-1.  NIech t=4x, stąd: [latex]hbox{tg}t=-1 \ t=-frac{pi}{4}+pi k \ 4x=-frac{pi}{4} + pi k \ x=-frac{pi}{16}+frac{1}{4} pi k[/latex] [latex]sin2x hbox{tg}x=sin x \ 2sin xcos x hbox{tg}x=sin x quad /:sin x \ 2 cos xhbox{tg}x=1 \ 2cos x frac{sin x}{cos x} = 1 \ 2sin x =1 \ sin x=frac{1}{2} \ x=frac{pi}{6}+2pi k quad hbox{oraz} quad x=frac{5pi}{6}+2 pi k[/latex] [latex]1+sin 2x=cos 2x \ underbrace{cos^{2}x+sin^{2}x}_{1}+sin2x=cos^{2}x-sin^{2}x \ sin^{2}x+2sin x cos x=-sin^{2}x\ sin x(sin x+2 cos x)=-sin^{2}x \ sin x+2cos x=-sin x \ 2sin x+ 2cos x=0 \ sin x + cos x=0 \ cos(frac{pi}{2}-x)+cos x=0 \ 2cosfrac{frac{pi}{2}-x+x}{2}cos frac{frac{pi}{2}-x-x}{2}=0 \ 2cosfrac{pi}{4}cos(frac{pi}{4}-x)=0 \ sqrt{2}cos(frac{pi}{4}-x)=0 \ cos(frac{pi}{4}-x)=0 [/latex] I ostatecznie: [latex] frac{pi}{4}-x=frac{pi}{2}+kpi \ -x=frac{pi}{4}+kpi \ x=-frac{pi}{4}-kpi[/latex]