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 = 02cos(3x + x)/2*cos(3x - x)/2 = 0cos2x * cosx = 0a)  cos2x = 02x = π/2 + πn, n∈Zx₁ = π/4 + πn/2, n ∈Zn = - 1x = π/4 - π/2 = - π/4 ∈ [- π/2;π/2]n = - 2x = π/4 - π = - 3π/4 ∉ [- π/2;π/2]n = 0x = 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/4b)  cosx = 0x = π/2 + πk, k∈Zk = - 1x = π/2 - π = - π/2 ∈ [- π/2;π/2]k = 0x = 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 = cos2x2sin²x - 2sinxcosx - (2cos²x - 1) = 02sin²x - 2sinxcosx - 2cos²x + sin²x + cos²x = 03sin²x - 2sinxcosx – cos²x = 0    / делим на cos²x ≠ 03tg²x - 2tgx - 1 = 0D = 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