Пошаговое объяснение:
ctg²x - tg²x = 4cos2x; 1/tg²x - tg²x = 4cos2x; ( (1 - tg²x) (1 + tg²x)) : tg²x = 4cos2x; ((1 - tg²x) · cos²x) : (cos²x · sin²x) = 4cos2x; (1 - sin²x/cos²x) : sin²x = 4cos2x; (cos²x - sin²x) : (cos²x · sin²x) = 4cos2x; ( 4cos2x) :(4cos²x · sin²x) = 4cos2x; cos2x (1/sin²x - 1) = 0; ОДЗ: sinx ≠ 0; cosx ≠ 0 ⇒cos2x = 0; 2x = π/2 + πn, n ∈ Z; X = π/4 + π/2n, n∈Z; ⇒ 1/ sin²2x = 1; sin²2x = 1 ⇒ sin2x = 1, sin2x = - 1; ⇒ 2x = π/2 + πn, n ∈ Z ; x = π/4 + π/2, n ∈ Z ответ: x = π/4 + π/2, n ∈ Z
