Решить уравнение Срочно сегодня до вечера

Question 1

Решить уравнение
Срочно сегодня до вечера

Question 2

$\frac{1}{cos^2x} + \frac{1}{cosx} -2=0\; ,\; \; \; ODZ:\; cosx\ne 0\; ,\; x\ne \frac{\pi}{2}+\pi n,\; n\in Z$

$t=cosx\; ,\; \; \frac{1}{t^2} +\frac{1}{t} -2=0\; ,\\\\ \frac{1+t-2t^2}{t^2}=0\; \; \to \; \; 2t^2-t-1=0\; ,\; \; D=9\; ,\\\\t_1= \frac{1-3}{4} =-\frac{1}{2}\; ,\; \; \; t_2= \frac{1+3}{4}=1\\\\1)\quad cosx=-\frac{1}{2}\; ,\; \; x=\pm arccos(-\frac{1}{2})+2\pi n,\\\\x=\pm (\pi -arccos\frac{1}{2})+2\pi n =\pm (\pi -\frac{\pi}{3})+2\pi n,\; n\in Z\\\\x_{1,2}=\pm \frac{2\pi}{3}+2\pi n,\; n\in Z$

$2)\quad cosx=1\; ,\; \; x_3=2\pi k,\; k\in Z\\\\3)\quad x\in [-5\pi ,-\frac{7\pi}{3}\, ]:\\\\a)\; n=-1:\; x_1= \frac{2\pi }{3}-2\pi =-\frac{4\pi}{3}\notin [-5\pi ,-\frac{7\pi }{3}]\\\\x_2=-\frac{2\pi}{3}-2\pi=-\frac{8\pi}{3}\in [-5\pi ,-\frac{7\pi}{3}]\\\\x_3=-2\pi \notin [-5\pi ,-\frac{7\pi}{3}]\\\\b)\; n=-2:\; x_1=\frac{2\pi}{3}-4\pi =-\frac{10\pi}{3}\in [-5\pi ,-\frac{7\pi}{3}]\\\\x_2=-\frac{2\pi}{3}-4\pi =-\frac{14\pi}{3}\in [-5\pi ,-\frac{7\pi}{3}]\\\\x_3=-4\pi$

$c)\; n=-3:\; x_1=\frac{2\pi}{3}-6\pi =-\frac{16\pi}{3}$ $\notin [-5\pi ,-\frac{7\pi}{3}]$

$x_2=-\frac{2\pi}{3}-6\pi =-\frac{20\pi}{3}\notin [-5\pi ,-\frac{7\pi}{3}]\\\\x_3=-6\pi \notin [-5\pi ,-\frac{7\pi}{3}]$

$Otvet:\ 1)x=\pm \frac{2\pi}{3}+2\pi n,\; n\in Z\; ;\; \; x=2\pi k,\; k\in Z\; ,\\\\2)\; -\frac{14\pi }{3} , \; -\frac{10\pi}{3} , \; -\frac{8\pi}{3} \; ,\; -4\pi \; .$