Решить уравнение 6sin^2+cosx-5=0 и найдите корни, принадлежащие отрезку [2 П, 3 П]

6 (1 - cos^2x) + cosx - 5 = 0 6 - 6cos^2x + cosx - 5 = 0 1 - 6cos^2x + cosx = 0 6cos^2x - cosx - 1 = 0 Пусть cosx = t 6t^2 - t - 1 = 0 D = 5^2 t1 = 1/2 t2 = 1/3 cosx = 1/2 x = + -pi/3 + 2pin cosx = 1/3 x = + - (pi - arccos1/3) + 2pin [2pi; 3pi] pi/3 + 2pin = > n = 1 = > 7pi/3 - pi/3 + 2pin = > - - (pi - arccos1/3) + 2pin = > n = 1 = > 3pi - arccos1/3 - (pi - arccos1/3) + 2pin = > - -