Cos 3x + cos x =0 промежуток [-п\2; п\2] и 2 sin в крадрате x - sin 2x=cos 2x заранее )

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 n = - 1 x =  π/4 -  π/2 = -  π/4  ∈ [-  π/2; π/2] n = - 2 x =  π/4 -  π = - 3π/4  ∉ [-  π/2; π/2] n = 0 x =  π/4  ∈ [-  π/2; π/2]n = 1x =  π/4 +  π/2 = 3π/4  ∉  [-  π/2; π/2] n = 2  π/4 +  π = 5π/4  ∉  [-  π/2; π/2]ответ: -  π/4;     π/4 b)   cosx = 0 x =  π/2 +  πk, k∈z k = - 1 x =  π/2 -  π = -  π/2  ∈  [-  π/2; π/2]k = 0 x =  π/2   ∈  [-  π/2; π/2]k = 1 x =  π/2 +  π = 3π/2  ∉    [-  π/2; π/2]ответ: -  π/2;   π/2  2)  2sin² x - sin2x = cos2x 2sin²x - 2sinxcosx - (2cos²x - 1) = 0 2sin²x - 2sinxcosx - 2cos²x + sin²x + cos²x = 03sin²x - 2sinxcosx – cos² x = 0     / делим на  cos²x  ≠ 0 3tg²x - 2tgx - 1 = 0 d = 4 + 4*3*1 = 16 1)  tgx = (2 - 4)/6 tgx = - 1/3x ₁ = - arctg(1/3) +  πn, n∈z tgx = ( 2 + 4)/6 tgx = 1  x₂ =  π/4 +  πk, k∈z

Log₃   xy=1   ⇒ху=3¹=