(4x+1)×(2x-4)-8x^2=3×(6-4)Найдите Вот это уравнение мне ​

1) cos2x  - 9cosx + 8 = 0
2cos²x - 1 - 9cosx + 8 = 0
2cos²x - 9cosx  + 7 = 0
D = 81 - 4*2*7 = 25
cosx = t, I t I  ≤ 1
2t² - 9t + 7 = 0
t₁ = (9 - 5)/4
t₁ = 1
t₂ = (9 + 5)/4
t₂ = 7/2 не удовлетворяет условию  I t I  ≤ 1
cosx = 1
x = 2πk, k∈Z 

2) 3cosx + sinx = 0  делим на cosx ≠ 0
3 + tgx = 0
tgx = - 3
x = - arctg(3) + πn, n∈Z

3) 3sin2x + sinxcosx - 2cos2x = 0
3sin2x + 1/2sin2x - 2cos2x = 0
 3,5* sin2x - 2cos2x = 0   делим на cos2x ≠ 0
3,5tg2x - 2 = 0
tg2x = 4/7
2x = arctg(4/7) + πn, n∈Z
x = (1/2)*arctg(4/7) + πn/2, n∈Z

1) sinx ≥  √2/2
 arcsin(√2/2) + 2πn ≤ x ≤ π - arcsin(√2/2) + 2πn, n∈Z
π/4 + 2πn ≤ x ≤ π - π/4 + 2πn, n∈Z
π/4 + 2πn ≤ x ≤  3π/4 + 2πn, n∈Z

2)  sinx ≥ 1/2
 arcsin(1/2) + 2πn ≤ x ≤ π - arcsin(1/2) + 2πn, n∈Z
π/3 + 2πn ≤ x ≤ π - π/3 + 2πn, n∈Z
π/3 + 2πn ≤ x ≤ 2π/3 + 2πn, n∈Z

3)  sinx< - √3/2
- π - arcsin(- √3/2) + 2πn < x < arcsin(- √3/2) +  2πn, n∈Z
-π + π/3 + 2πn < x < - π/3 + 2πn, n∈Z
-2π/3  + 2πn < x < - π/3 + 2πn, n∈Z

4)  sinx  < -(√2/2)
- π - arcsin(- √2/2) + 2πn < x < arcsin(- √2/2) +  2πn, n∈Z
- π + π/4 + 2πn < x < - π/4 + 2πn, n∈Z
- 3π/4 + 2πn < x < - π/4 + 2πn, n∈Z