Решите тригометрическое уравнение: a) 2sin2x-1=0 b) cos^2x -cos x =0 буду ; )

а) 2sin2x-1 = 0, sin2x = 1/2

2x = (-1)^n*arcsin1/2 + πn

2x = (-1)^n *π/6 + πn

x =  (-1)^n* π/12 + πn/2

б) cos²x - cosx = 0

cosx(cosx-1) = 0

б.1) cosx = 0    

x = π/2 + πn

б.2) cosx -1 = 0

cosx = 1

x = 2πn