Решите уравнение: 4cos²x - 2sin²x - 5cos x - 4 = 0

Решите задачу:

${sin}^{2} x = 1 - {cos}^{2} x \\ 4 {cos}^{2} x - 2(1 - {cos}^{2} x) - 5cosx - 4 = 0 \\ 4 {cos}^{2} x - 2 + 2 {cos}^{2} x - 5cosx - 4 = 0 \\ 6 {cos}^{2} x - 5cosx - 6 = 0 \\ cosx = t \\ (- 1 \leqslant t \leqslant 1 )\\ 6 {t}^{2} - 5t - 6 = 0 \\ d = {b}^{2} - 4ac = 25 - 4 \times 6 \times ( - 6) = 169 \\ t1 = \frac{5 + 13}{12} = \frac{18}{12} = \frac{3}{2} \\ t2 = \frac{5 - 13}{12} = \frac{ - 8}{12} = - \frac{2}{3} \\ cosx = - \frac{2}{3} \\ x = + - arccos( - \frac{2}{3} ) + 2\pi n$