2sin^2x+sinx=1 ; ^2- квадрат 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