Решите уравнение 2sin^2 x-cos x-1=0

2*(1 - cos^2(x)) - cosx - 1 = 0 2 - 2cos^2(x) - cos(x) - 1 = 0 -2cos^2(x) - cos(x) + 3 = 0 2cos^2(x) + cos(x) - 3 = 0 Замена: cos(x) = t, -1 ≤ t ≤ 1 2t^2 + t - 3 = 0 D = 1 + 4*3*2 = 1 + 24 = 25 t1 = (-1 - 5)/4 = -6/4 = -3/2 < -1 - посторонний корень t2 = (-1 + 5)/4 = 4/4 = 1 cosx = 1, x = 2πk