Question

tgx+6ctg-5=0 помогите плиз

Answer 1

$tgx+6ctgx-5=0 \\ \frac{1}{ctgx} +6ctgx-5=0 | *ctgx \\ 1+6 ctg^{2} x-5ctgx=0 \\ ctgx=t, t>0 \\ 6 t^{2} -5t+1=0 \\ D=5*5-4*6=25-24=1 \\ t_{1} = \frac{5+1}{12} = \frac{6}{12} = \frac{1}{2} \\ t_{2} = \frac{5-1}{12} = \frac{4}{12} = \frac{1}{3} \\ ctgx= \frac{1}{2} \\ x_{1} =arcctg \frac{1}{2}+\pi n, n E z \\ ctgx= \frac{1}{3} \\ x_{2} =arcctg \frac{1}{3}+\pi n, n E z$