Решите тригонометрические неравенства ( + киньте формулы, которые необходимо знать для...

Question

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

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Answer 1

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<br>-π/10+πn/5x∈(-π/10+πn/5;-π/15+πn/5,n∈z)
3
x∈(5π/6+2πn:13π/6+2πn,n∈z)
4
x∈[π/4+πn;π/2+πn,n∈z)
5
sinx=a
6a²-8a+2,5<0<br>D=64-60=4
a1=(8-2)/12=1/2
a2=(8+2)/12=5/6
1/2sinx>1/2⇒x∈(π/6+2πn;5π/6+2πn,n∈z)
sinx<5/6⇒x∈(arcsin5/6+2πn;π-arcsin5/6+2πn,n∈z<br>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<br>πn/2x∈(πn/2;π/8+πn/2,n∈z)
7
√3/cos²x-4sinx/cosx<0<br>(√3-4sinxcosx)/cos²x<0<br>cos²x>0  ⇒√3-2sin2x<0<br>sin2x>√3/2
π/3+2πn<2x<2π/3+2πn,n∈z<br>π/6+πnx∈(π/6+πn;π/3+πn,n∈z)

Answer 2

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<br>- 7π/6 + 2πn < x < π/6 + 2πn, n ∈ Z<br>4)  tgx ≥ 1
arctg(1) + πm ≤ x < π/2  + πm, m ∈ Z<br> π/4 + πm ≤ x < π/2  + πm, m ∈ Z<br>
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<br>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<br>kπ/2 < x < π/8 + πk/2, k ∈ Z<br>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<br>