Решите тригонометрические неравенства ( + киньте формулы, которые необходимо знать для решения подобных заданий)

Решение1)  sin2x ≥ √3/2arcsin(√3/2) + 2πn ≤ 2x ≤ (π - arcsin(√3/2)) + 2πn, n ∈ Zπ/3 + 2πn ≤ 2x ≤ (π - π/3) + 2πn, n ∈ Zπ/3 + 2πn ≤ 2x ≤ 2π/3 + 2πn, n ∈ Zπ/6 + πn ≤ x ≤ π/3 + πn, n ∈ Z2)  tg5x < - √3πk - π/2 < 5x < arctg(- √3) + πk, k ∈ Zπk - π/2 < 5x < - π/3 + πk, k ∈ Zπk/5 - π/10 < x < - π/15 + πk/5, k ∈ Z3)  sinx < 1/2- π - arcsin(1/2) + 2πn < x < arcsin(1/2) + 2πn, n ∈ Z- π - π/6 + 2πn < x < π/6 + 2πn, n ∈ Z- 7π/6 + 2πn < x < π/6 + 2πn, n ∈ Z4)  tgx ≥ 1arctg(1) + πm ≤ x < π/2  + πm, m ∈ Z π/4 + πm ≤ x < π/2  + πm, m ∈ Z5)  6sin²x - 8sinx + 2,5 < 0 sinx = t6t² - 8t + 2,5 = 0D = 64 - 4*6*2,5 = 4t₁ = (8 - 2)/12t₁ = 1/2t₂ = (8 + 2)/12t₂ = 5/61/2 < sinx < 5/6а)  sinx > 1/2arcsin(1/2) + 2πk < x < (π - arcsin(1/2)) + 2πk, k ∈ Zπ/6 + 2πk < x < (π - π/6) + 2πk, k ∈ Zπ/6 + 2πk< x < 5π/6 + 2πk, k ∈ Z б)  sinx < 5/6- π - arcsin(5/6) + 2πk < x < arcsin(5/6) + 2πk, k ∈ ZОтвет: x ∈ (π/6 + 2πk ; arcsin(5/6)+ 2πk, k ∈ Z)- π - arcsin(5/6) + 2πk ; 5π/6 + 2πk, k ∈ Z 6)  sin4x + cos4x * ctg2x > 12sin2x*cos2x + {[(1 - 2sin²2x)*co2x] / sin2x} - 1 > 0(2sin²2x * cos2x + cos2x - 2sin²2x * cos2x - sin2x) / sin2x  > 0(cos2x - sin2x)/sin2x > 0ctg2x - 1 > 0ctg2x > 1kπ < 2x < arcctg1 + πk, k ∈ Zkπ < 2x < π/4 + πk, k ∈ Zkπ/2 < x < π/8 + πk/2, k ∈ Z7)  √3 / cos²x < 4tgx(4tgx * cos²x - √3) / cos²x > 0(2sin2x - √3)/cos²x > 0cos²x ≠ 0, x ≠ π/2 + πn, n ∈ Z2sin2x - √3 > 0sin2x > √3/2arcsin(√3/2) + 2πk < 2x < (π - arcsin(√3/2)) + 2πk, k ∈ Zπ/3 + 2πk < 2x < (π - π/3) + 2πk, k ∈ Zπ/3 + 2πk < 2x < 2π/3 + 2πk, k ∈ zπ/6 + πk < x < π/3 + πk, k ∈ Z