1) Cos(x+P/6)-1=0 2)SinX/4CosX/4=-1/4 3)4Sin^x-4CosX-1=0 Пожалуйста помогите с...

Решите задачу:

$1)\ cos(x + \dfrac{ \pi }{6}) - 1 = 0 \\ \\ cos(x + \dfrac{ \pi }{6}) = 1 \\ \\ x + \dfrac{ \pi }{6} = 2 \pi n, \ n \in Z \\ \\ x = -\dfrac{ \pi }{6} + 2 \pi n, \ n \in Z$

$2) \ sin \dfrac{x}{4} cos \dfrac{x}{4} = - \dfrac{1}{4} \\ \\ 2sin \dfrac{x}{4} cos \dfrac{x}{4} = -\dfrac{1}{2} \\ \\ sin\dfrac{x}{2} = -\dfrac{1}{2} \\ \\ \dfrac{x}{2} = (-1)^{n + 1} \dfrac{ \pi }{6} + \pi n, \ n \in Z \\ \\ x = (-1)^{n + 1} \dfrac{ \pi }{3} + 2\pi n, \ n \in Z$

$3) \ sin^2x - cosx - 1 = 0 \\ \\ 1 - cos^2x - cosx - 1 = 0 \\ \\ -cos^2x - cosx = 0 \\ \\ cos^2x + cosx = 0 \\ \\ cosx(cosx + 1) = 0 \\ \\ cosx = 0 \\ \\ \boxed{x = \dfrac{ \pi }{2} + \pi n, \ n \in Z } \\ \\ cosx = -1 \\ \\ \boxed{x = \pi + 2 \pi k, \ k \in Z}$