Ответ(ы) на вопрос:
Гость1)Sinx > √3/2
π/3 + 2πk < x < 2π/3 + 2πk , k ∈Z
2) Ctgx < -1
3π/4 + πk < π + πk , k∈Z
3) 4Sin²x -11Cosx -1 = 0
4(1-Cos²x) -11Cosx -1 = 0
4 -4Cos²x -11Cosx -1 = 0
-4Cos²x -11Cosx +3 = 0
4Cos²x +11Cosx -3 = 0
Решаем как квадратное
D = b² -4ac = 121 - 4*4*(-3) = 169
Cosx = (-11+13)/8 = 1/4 Сosx = (-11 -13)/8= -3
x = +-arcCos(1/4) = 2πk , k ∈Z ∅
4) 3Sin²x - 2SinxCosx - Cos²x = 2Sin²x + 2Cos²x
Sin²x -2SinxCosx -3Cos²x = 0 | : Cos²x ≠ 0
tg²x -2tgx -3 = 0
решаем как квадратное по т. Виета
tgx = 3 tg x = -1
x = arctg3 + πk ,k ∈Z x = -π/4 + πn , n ∈Z
5) Cosx - Co3x +Sinx = 0
2Sin2xSinx + Sinx = 0
Sinx(2Sin2x +1) = 0
Sinx = 0 или 2Sin 2x +1 = 0
x = πn, n ∈ Z 2Sin2x = -1
Sin2x = -1/2
2x = (-1)ⁿarcSin(-1/2) + nπ, n ∈Z
2x = (-1)ⁿ * (-π/6) + nπ, n ∈Z
x = (-1)ⁿ+₁ π/6 + nπ, n ∈Z
6)Sin3x -Cos3x = √2 Sinx
Sin3x - Sin(π/2 - 3x) = √2Sinx
2Cosπ/4Sin(3x -π/4) = √2Sinx
√2 Sin(3x -π/4) =√2Sinx
Sin(3x - π/4) - Sinx = 0
2Cos(2x - π/8) Sin(x-π/8) = 0
Cos(2x - π/8) = 0 или Sin(x-π/8) = 0
2х - π/8 = π/2 + πк, к ∈Z х - π/8 = πn , n∈Z
x = π/16 +π/4 + πк/2, к ∈Z x = π/8 + πn, n ∈Z
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