2sin^2x+sinx=1; ^2 - квадрат

Помогите решить

2sin^2x + sinx = 1

2sin^2x + sinx - 1 = 0

sinx = t

2t^2 + t - 1 = 0

D = 1 + 4*2*1 = 9

t1 = (-1 - 3) / 4

t1 = - 1

t2 = (-1 + 3) / 4

t2 = 1/2

1) sinx = - 1

x1 = - π/2 + 2πn, n∈Z

2) sinx = 1/2

x = (-1) ^n*arcsin (1/2) + πk, k∈Z

x2 = (-1) ^n * (π/6) + πk, k∈Z