Помогите решить пример по алгебре.

Решите задачу:

$\frac{6}{tg^2x} + \frac{5}{tgx} - 1 = 0; [-\pi; \frac{\pi}{2}]\\\\tgx=t; cosx\neq0;t\neq\frac{\pi}{2}\\\\\frac{6}{t^2} + \frac{5}{t} - 1 = 0\\\\\frac{6+5t-6t^2}{t^2} = 0\\\\t^2\neq0\\\\-6t^2+5t+6=0\\6t^2-5t-6=0\\D=25+144=169\\t_1=\frac{5+13}{12}=\frac{3}{2}\\t_2=\frac{5-13}{12}=-\frac{2}{3}\\\\tgx=\frac{3}{2}\\x=arctg\frac{3}{2}+\pi*k, k\in Z\\\\tgx=-arctg\frac{2}{3}+\pi*k, k\in Z\\\\-\pi \leqarctg\frac{3}{2}+\pi*k\leq\frac{\pi}{2}\\-\pi-arctg\frac{3}{2}\leq\pi*k\leq\frac{\pi}{2}-arctg\frac{3}{2}$

$\frac{-\pi-arctg\frac{3}{2}}{\pi}\leq k\leq\frac{\frac{\pi}{2}-arctg\frac{3}{2}}{\pi}\\\\-\pi\leq-arctg\frac{2}{3}+\pi*k\leq \frac{\pi}{2}\\-pi+arctg\frac{2}{3}\leq \pi*k\leq \frac{\pi}{2}+arctg\frac{2}{3}\\\frac{-\pi+arctg\frac{2}{3}}{\pi}\leq k\leq\frac{\frac{\pi}{2}+arctg\frac{2}{3}}{\pi}$