Решите уравнение cos2x+2cos^2x-sin2x=0 и найдите корни, принадлежащие отрезку [3π/2; 5π/2]

a) cos2x + 2cos^2x - sin2x = 0 cos^2x - sin^2x + 2cos^2x - 2sinxcosx = 0 3cos^2x - sin^2x - 2sinxcosx = 0 | : cos^2x 3 - tg^2x - 2tgx = 0 Пусть tgx = t 3 - t^2 - 2t = 0 t^2 + 2t - 3 = 0 Д = 4 + 12 = 16 = 4^2 t1 = 1; t2 = - 3 tgx = 1 x = pi/4 + pin; tgx = - 3 x = - arctg3 + pin; б) pi/4 + pin = > n = 2 9pi/4; - arctg3 + pin = > n = 2 - arctg3 + 2pin Ответ: 9pi/4; - arctg3 + 2pin.