решите уравнение: (8cos^2x+6cosx-5)*log_7(-sinx)=0

$(8cos^2x+ 6cosx - 5)log_7(-sinx) = 0\\\\ sin x < 0; x \in (\pi + 2\pi n; 2\pi + 2\pi n), \ n \in Z\\\\ 1) \ 8cos^2x + 6cosx - 5 = 0\\\\ cosx = t, -1 \leq t \leq 1\\\\ 8t^2 + 6t - 5 = 0\\\\ D = 36 + 190 = 196\\\\ t_1 = \frac{-6 - 14}{16} < -1, \ t_2 = \frac{-6 + 14}{16} = \underline{ \frac{1}{2} }\\\\ cosx = \frac{1}{2}\\\\ x = \frac{\pi}{3} + 2\pi n, \ n \in Z\\\\ \boxed{ x = -\frac{\pi}{3} + 2\pi n, \ n \in Z}\\\\ 2) \ log_7(-sinx) = log_71\\\\ -sinx = 1\\\\ \boxed{x = -\frac{\pi}{2} + 2\pi n, \ n \in Z}$