Тригонометрические уравнения..... .

Решите задачу:

$1)\; \; 6cos^4x-11cos^2x+4=0\\\\t=cos^2x\; ,\; \; 0 \leq cos^2x \leq 1\; \; \to \; 0 \leq t\leq 1\\\\6t^2-11t+4=0\; ,\; \; D=25\; ,\\\\t_1= \frac{11-5}{12}=\frac{1}{2}\ \textless \ 1\; ,\; \; t_2= \frac{11+5}{12}= \frac{4}{3} \ \textgreater \ 1\\\\cos^2x= \frac{1}{2}\; \; \Rightarrow \; \; \frac{1+cos2x}{2} = \frac{1}{2} \; \; \Rightarrow \; \; cos2x=0\\\\2x=\frac{\pi}{2}+\pi n\; ,\; \; \; x=\frac{\pi}{4}+\frac{\pi n}{2}\; ,\; n\in Z\\\\x\in [-\frac{7\pi}{2};-2\pi ]:\; \; x=- \frac{9\pi }{4} \; ,\; -\frac{11\pi }{4}\; ,\; \; -\frac{13\pi }{4}\; .$

$2)\; \; (2cos^2x-cosx)\sqrt{-11tgx}=0\; ,\; \; ODZ:\; \; x\ne \frac{\pi }{2}+\pi n,\; n\in Z\\\\ \left \{ {{cosx(2cosx-1)=0} \atop {tgx \leq 0\; ,\; x\ne \frac{\pi}{2}+\pi n}} \right. \; \; \left \{ {{cosx=0\; \; ili\; \; cosx=\frac{1}{2}} \atop {-\frac{\pi}{2}+\pi n\ \textless \ x \leq \pi n\; ,\; x\ne \frac{\pi}{2}+\pi n}} \right. \\\\ \left \{ {{x=\pm \frac{\pi}{3}+2\pi n} \atop {-\frac{\pi}{2}+\pi n\ \textless \ x\leq \pi n}} \right. \; \; \to \; \; x=-\frac{\pi}{3}+2\pi n\; ,\; \; x=2\pi n\; ,\; n\in Z\\\\Otvet:\; \; x=- \frac{\pi }{3}+2\pi n\; ,\; \; x=2\pi n\; ,\; n\in Z\; .$