5cos²x+7cosx-6=08cos²x-10sinx-11=0

Решите задачу:

$5cos^2x+7cosx-6=0\\\\cosx=t,\; \; 5t^2+7t-6=0\\\\D=49+4*5*6=169\\\\t_1=\frac{-7-13}{10}=-2,\; \; t_2=\frac{6}{10}=\frac{3}{5}\\\\cosx=-2\ \textless \ -1\; \to \; net\; resheniya\\\\cosx=\frac{3}{5},\; x=\pm arccos\frac{3}{5}+2\pi n,\; n\in Z$

$8cos^2x-10sinx-11=0\\\\8(1-sin^2x)-10sinx-11=0\\\\8sin^2x+10sinx+2=0\\\\t=sinx,\; 8t^2+10t+2=0;4t^2+5t+1=0\\\\D=25-16=9\\\\t_1=\frac{-5-3}{8}=-1,\; \; t_2=\frac{-5+3}{8}=-\frac{1}{4}\\\\sinx=-1,\; \; x=-\frac{\pi}{2}+2\pi n,\; n\in Z\\\\sinx=-\frac{1}{4},\\\\ x=(-1)^{k}\cdot arcsin(-\frac{1}{4})+\pi k=(-1)^{k+1}\cdot arccos\frac{1}{4}+\pi k,\; k\in Z$