Cos 3x + cos x =0 промежуток [-П\2;П\2] и 2 sin в крадрате x - sin 2x=cos 2x Заранее спасибо)

1)  cos3x + cosx = 0 2cos(3x + x)/2*cos(3x - x)/2 = 0 cos2x * cosx = 0 a)  cos2x = 0 2x = π/2 + πn, n∈Z x₁ = π/4 + πn/2, n ∈Z b)  cosx = 0 x = π/2 + πk, k∈Z 2) 2sin² x - sin2x = cos2x 2sin²x - 2sinxcosx - (2cos²x - 1) = 0 2sin²x - 2sinxcosx - 2cos²x + sin²x + cos²x = 0 3sin²x - 2sinxcosx – cos²x = 0    / делим на cos²x ≠ 0 3tg²x - 2tgx - 1 = 0 D = 4 + 4*3*1 = 161) tgx = (2 - 4)/6tgx = - 1/3x₁ = - arctg(1/3) + πn, n∈Ztgx = ( 2 + 4)/6tgx = 1 x₂ = π/4 + πk, k∈Z