6sin квадрат x + 5 cos x - 7=0

$6sin^2x + 5cosx - 7 = 0 \\ 6sin^2x - 6 + 5cosx - 1 = 0 \\ -6cos^2x + 5cosx - 1 = 0 \\ 6cos^2x - 5cosx + 1 = 0$
Пусть $t = cosx, \ t \in [-1; 1].$
$6t^2 - 5t + 1 = 0 \\ D = 25 - 4 \cdot 6 = 1 \\ \\ t_1 = \dfrac{5 + 1}{12} = \dfrac{1}{2} \\ \\ t_2 = \dfrac{5-1}{12} = \dfrac{1}{3}$
Обратная замена:
$cosx = \dfrac{1}{2} \\ \\ \boxed{x= \pm \frac{ \pi }{3} +2 \pi n, \ n \in Z} \\ \\ cosx = \dfrac{1}{3} \\ \\ \boxed{x = \pm arccos \dfrac{1}{3} + 2 \pi n, \ n \in Z.}$