Ответ:
1) x₁ = -5π/6 + 20 + 10πk, k ∈ Z
x₂ = 35π/6 + 20 + 10πk, k ∈ Z
или
2) x = -5π/6 ⋅ (-1)ᵏ + 20 + 5πk, k ∈ Z
Пошаговое объяснение:
2sin(x/5 - 4) + 1 = 0 | -1
2sin(x/5 - 4) = -1 | :2
sin(x/5 - 4) = -1/2
Решить уравнение sin t = a можно двумя способами:
1) x₁/5 - 4 = arcsin(-1/2) + 2πk = -π/6 + 2πk | +4
x₁/5 = -π/6 + 4 + 2πk | ⋅ 5
x₁ = -5π/6 + 20 + 10πk, k ∈ Z
x₂/5 - 4 = π - arcsin (-1/2) + 2πk = π - (-π/6) + 2πk = 7π/6 + 2πk | +4
x₂/5 = 7π/6 + 4 + 2πk | ⋅ 5
x₂ = 35π/6 + 20 + 10πk, k ∈ Z
2) x/5 - 4 = (-1)ᵏ ⋅ arcsin (-1/2) + πk = (-1)ᵏ ⋅ (-π/6) + πk | +4
x/5 = (-1)ᵏ ⋅ (-π/6) + 4 + πk | ⋅ 5
x = -5π/6 ⋅ (-1)ᵏ + 20 + 5πk, k ∈ Z