Помогите, пожалуйста, с решением.
2*cos^2(x)=8cos2x-sin2x

2Cos²x = 8Cos2x - Sin2x2Cos²x - 8(Cos²x - Sin²x) + 2SinxCosx = 02Cos²x - 8Cos²x  + 8Sin²x + 2SinxCosx = 0-6Cos²x +8 Sin²x + 2SinxCosx = 03Cos²x - 4Sin²x - SinxCosx =0 | : Cos²x≠ 03 -4tg²x -tgx = 0tgx = y4y² + y -3 = 0D = 49y₁ = -1+7)/8 = 6/8 = 3/4у₂ = (-1 -7)/8 = -1а) у = 3/4tgx = 3/4x = arctg3/4 + πk , k ∈ Zб) y = -1tgx = -1x = -π/4 + πn , n ∈Z