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