Question

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

Answer 1

 

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