2sinxcos2x-sinx+2cos2x-1=0

Помогите решить

2SinxCos2x-Sinx+2Cos2x-1=0

2Cos2x (Sinx + 1) - (Sinx + 1) = 0

(Sinx + 1) (2Cos2x - 1) = 0

Sinx + 1 = 0 или 2 Сos2x - 1 = 0

Sinx = - 1 Cos2x = 1/2

x = - π/2 + 2πk, k ∈Z 2x = + - arcCos1/2 + 2πn, n ∈Z

2x = + - π/3 + 2πn, n ∈Z

x = + - π/6 + πn, n ∈ Z