Помогите решить)sin²α-cos²α=cosα

Решите задачу:

$sin ^{2} \alpha -cos ^{2} \alpha =cos \alpha \\ (1-cos ^{2} \alpha )-cos ^{2} \alpha -cos \alpha =0 \\ 1-cos ^{2} \alpha -cos ^{2} \alpha -cos \alpha =0 \\ 1-2cos ^{2} \alpha -cos \alpha =0 /*(-1)\\ 2cos ^{2} \alpha +cos \alpha -1=0 \\ cosx=t \\ 2t ^{2} +t-1=0 \\ D=1+8=9 \\ \sqrt{D} =3 \\ t _{1} = \frac{-1+3}{4} = \frac{2}{4} = \frac{1}{2} \\ t _{2} = \frac{-1-3}{4} =-1 \\ cosx= \frac{1}{2} \\ x=+- \frac{ \pi }{3} + 2\pi n \\$ $cosx=-1 \\ x= \pi +2 \pi n$