Question

Решите уравнение
1+соs x=2sin^2 x

Answer 1

$1 + cosx = 2sin^2x \ \ 1 + cosx = 2 - 2cos^2x \\ 2cos^2x + cosx - 1 = 0$
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:
$cosx = dfrac{1}{2} \ \ boxed{x = pm dfrac{ pi }{3} + 2 pi n, n in Z }\ \ cosx = -1 \ \ boxed{x = pi + 2 pi n, n in Z}$