<span>y=x^2 +3x+2.
1. y(-2)=4-6+2=0
2. </span>x^2 +3x+2=6 x^2 +3x-4=0 x1=-4 x2=1 (по т. Виета)
3. x^2 +3x+2=0 x1=-2 x2=-1 (по т. Виета)
x^2 - (x^2 - 1^2) = x^2 - x^2 + 1 = 1
Фармула дискриминантаD=b^2-4ас
D=25-16=9>0…значит 2 корня=
сosB=sinA=10/109(ko..)
sinA=(ko..)1-cosA=(ko..1-(10/109)^2)=7/(ko..109)
tgA=sinA/cosA=(7/ko..109)/(10/ko..109)=7/10
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