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

Решение 1)  sin2x ≥ √3/2 arcsin(√3/2) + 2πn ≤ 2x ≤ (π - arcsin(√3/2)) + 2πn, n ∈ Z π/3 + 2πn ≤ 2x ≤ (π - π/3) + 2πn, n ∈ Z π/3 + 2πn ≤ 2x ≤ 2π/3 + 2πn, n ∈ Z π/6 + πn ≤ x ≤ π/3 + πn, n ∈ Z 2)  tg5x < - √3 πk - π/2 < 5x < arctg(- √3) + πk, k ∈ Z πk - π/2 < 5x < - π/3 + πk, k ∈ Z πk/5 - π/10 < x < - π/15 + πk/5, k ∈ Z 3)  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 ∈ Z 4)  tgx ≥ 1 arctg(1) + πm ≤ x < π/2  + πm, m ∈ Z  π/4 + πm ≤ x < π/2  + πm, m ∈ Z 5)  6sin²x - 8sinx + 2,5 < 0  sinx = t 6t² - 8t + 2,5 = 0 D = 64 - 4*6*2,5 = 4 t₁ = (8 - 2)/12 t₁ = 1/2 t₂ = (8 + 2)/12 t₂ = 5/6 1/2 < sinx < 5/6 а)  sinx > 1/2 arcsin(1/2) + 2πk < x < (π - arcsin(1/2)) + 2πk, k ∈ Z π/6 + 2πk < x < (π - π/6) + 2πk, k ∈ Z π/6 + 2πk< x < 5π/6 + 2πk, k ∈ Z  б)  sinx < 5/6 - π - arcsin(5/6) + 2πk < x < arcsin(5/6) + 2πk, k ∈ Z Ответ: x ∈ (π/6 + 2πk ; arcsin(5/6)+ 2πk, k ∈ Z) - π - arcsin(5/6) + 2πk ; 5π/6 + 2πk, k ∈ Z  6)  sin4x + cos4x * ctg2x > 1 2sin2x*cos2x + {[(1 - 2sin²2x)*co2x] / sin2x} - 1 > 0 (2sin²2x * cos2x + cos2x - 2sin²2x * cos2x - sin2x) / sin2x  > 0 (cos2x - sin2x)/sin2x > 0 ctg2x - 1 > 0 ctg2x > 1 kπ < 2x < arcctg1 + πk, k ∈ Z kπ < 2x < π/4 + πk, k ∈ Z kπ/2 < x < π/8 + πk/2, k ∈ Z 7)  √3 / cos²x < 4tgx (4tgx * cos²x - √3) / cos²x > 0 (2sin2x - √3)/cos²x > 0 cos²x ≠ 0, x ≠ π/2 + πn, n ∈ Z 2sin2x - √3 > 0 sin2x > √3/2 arcsin(√3/2) + 2πk < 2x < (π - arcsin(√3/2)) + 2πk, k ∈ Z π/3 + 2πk < 2x < (π - π/3) + 2πk, k ∈ Z π/3 + 2πk < 2x < 2π/3 + 2πk, k ∈ z π/6 + πk < x < π/3 + πk, k ∈ Z

1 π/3+2πn≤2x≤2π/3+2πn,n∈z π/6+πn≤x≤π/3+πn,n∈z x∈[π/6+πn;π/3+πn,n∈z] 2 -π/2+πn<5x<-π/3+πn,n∈z -π/10+πn/51/2⇒x∈(π/6+2πn;5π/6+2πn,n∈z) sinx<5/6⇒x∈(arcsin5/6+2πn;π-arcsin5/6+2πn,n∈z Ответ x∈(π/6+2πn;arcsin5/6+2πn,n∈z) U (5π/6+2πn;π-arcsin5/6+2πn,n∈z) 6 sin4x+cos4xcos2x/sin2x>1 (sin4xsin2x+cos4xcos2x)/sin2x>1 cos2x/sin2x>1 ctg2x>1 πn<2x<π/4+πn,n∈z πn/20  ⇒√3-2sin2x<0 sin2x>√3/2 π/3+2πn<2x<2π/3+2πn,n∈z π/6+πn