Question

1/cos^2 3x-7tg3x+11=0

Answer 1

$\displaystyle\bf\\\frac{1}{Cos^{2}3x}-7tg3x+11=0\\\\\\tg^{2} 3x+1-7tg3x+11=0\\\\tg^{2} 3x-7tg3x+12=0\\\\tg3x=m\\\\m^{2} -7m+12=0\\\\D=(-7)^{2} -4\cdot 12=49-48=1\\\\\\m_{1} =\frac{7-1}{2} =3\\\\\\m_{2} =\frac{7+1}{2} =4\\\\1) \ tg3x=3\\\\3x=arctg3+\pi n,n\in Z\\\\x_{1} =\frac{1}{3} arctg3+\frac{\pi n}{3} ,n\in Z\\\\2) \ tg3x=4\\\\3x=arctg4+\pi n,n\in Z\\\\x_{2} =\frac{1}{3} arctg4+\frac{\pi n}{3} ,n\in Z$