6sin^2x - 5sinx-4=0 [-7 пи /2 ; -3 пи/2]

Решите задачу:

$6sin^2x - 5sinx - 4 = 0 \\ \\ t = sinx, \ t \in [-1; \ 1] \\ \\ 6t^2 - 5t - 4 = 0 \\ \\ D = 25 + 4 \cdot 4 \cdot 6 = 96 + 25 = 121 = 11^2 \\ \\ t_1 = \dfrac{5 + 11}{12} = \dfrac{16}{12} - \ an \ extraneous \ root \\ \\ t_2 = \dfrac{5 - 11}{12} = -\dfrac{1}{2} \\ \\ Reverse \ \ substitution: \\ \\ sinx = -\dfrac{1}{2} \\ \\ x = (-1)^{n+1} \dfrac{\pi} {6} + \pi n, \ n \in Z \\ \\ - \dfrac{7 \pi }{2} \leq (-1)^{n+1} \dfrac{\pi} {6} + \pi n \leq - \dfrac{3 \pi }{2}, \ n \in Z$
$-21 \pi \leq (-1)^{n + 1} \pi + 6 \pi n \leq -9 \pi , \ n \in Z \\ \\ -21 \leq (-1)^{n + 1} + 6n \leq -9 , \ n \in Z \\ \\ n = -3; -2 \\ \\ x_1 = (-1)^{-3 + 1} \dfrac{\pi }{6} - 3 \pi = (-1)^{-2} \dfrac{ \pi} {6} - 3 \pi = \dfrac{\pi }{6} - 3 \pi = - \dfrac{17 \pi }{6} \\ \\ x_2 = (-1)^{-2 + 1}\dfrac{\pi }{6} - 2 \pi = - \dfrac{\pi }{6} - 2 \pi = - \dfrac{13 \pi }{6} \\ \\ \boxed {Answer: \ x = - \dfrac{17 \pi }{6}; \ - \dfrac{13 \pi }{6}.}$