Question

помогите решить)
sin²α-cos²α=cosα

Answer 1

$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} + 2pi n$$cosx=-1 \ x= pi +2 pi n$