Решите уравнение tgx=√3,cosx=2; ctgx=√3

1. sin(3x)=1/2; 3x=(-1)^k*pi/6 +pik x=(-1)^k*pi/18 +pik/3, 2. cos(x/2)=-sgrt3/2; x/2=+-5pi/6+2pik x=+-5pi/3+4pik/ 3. ctg(x-pi/4)=sgrt3; x-pi/4=pi/6+pik; x=pi/6+pi/4+pik; x=5pi/12 +pik. 4. 2cos^2x-cosx-1=0 cosx=t; 2t^2-t-1=0 t1=1: cosx=1; x=2pik; t2=-1/2; cosx=-1/2; x=+-pi/3+2pik/ 5. 3tgx-2/tgx - 1=0 3tg^2x-tgx-2=0 tgx=t 3t^2-t-2=0 t1=1; tgx=1; x=pi/4+pik t2=-2/3; tgx=-2/3; x=-arctg(2/3)+pik/ 6. 1-2sin^2(x/3) +5sin(x/3)+2=0; 2sin^2(x/3)-5sin(x/3)-3=0 sin(x/3)=t; 2t^2-5t-3=0 t1=-1; sin(x/3)=-1; x/3=-pi/2+2pik; x=-3pi/2+6pik=pi/2+6pik; t2=3 >1 Нет решений. ОТвет:x=pi/2+6pik