Решите уравнение 1+соs x=2sin^2 x

$1 + cosx = 2sin^2x \\ \\ 1 + cosx = 2 - 2cos^2x \\\\ 2cos^2x + cosx - 1 = 0$
Пусть $t = cosx, \ |t| \leq 1$
$2t^2 + t - 1 = 0 \\ \\ D = 1 + 8 = 9 = 3^2 \\ \\ t_1 = \dfrac{-1 + 3}{4} = \dfrac{1}{2} \\ \\ t_2 = \dfrac{-1 - 3}{4} = -1$
Обратная замена:
$cosx = \dfrac{1}{2} \\ \\ \boxed{x = \pm \dfrac{ \pi }{3} + 2 \pi n, \ n \in Z }\\ \\ cosx = -1 \\ \\ \boxed{x = \pi + 2 \pi n, \ n \in Z}$