Помогите решить уравнение:

$\sin^{2} x + \cos^{2}x = 1 \\ { \sin }^{2} x = 1 - { \cos}^{2} x \\ 2 \times (1 - \cos^{2} x) - 5 = - 5 cosx \\ 2 - 2 cos^{2} x - 5 = - 5cosx \\ 2 cos^{2} x - 5cosx + 3 = 0 \\cosx = t \\ 2 {t}^{2} - 5t + 3 = 0 \\ d = {b}^{2} - 4ac = 25 - 4 \times 2 \times 3 = 1 \\ t1 = \frac{5 + 1}{2 \times 2} = \frac{3}{2} \\ t2 = \frac{5 - 1}{2 \times 2} = 1 \\ cosx = 1 \\ x = 2\pi n$
Ответ: с)