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