Question

Решите уравнение sin(2х+5π/6)=√3/2 на промежутке (0;2π).

Answer 1

Решение:

$sin(2x+\frac{5\pi}{6}) = \frac{\sqrt{3}}{2} \\2x+\frac{5\pi}{6} = (-1)^{n}*arcsin(\frac{\sqrt{3}}{2}) + \pi n \\2x+\frac{5\pi}{6} = (-1)^{n}*\frac{\pi}{3}+\pi n \\2x = (-1)^{n}*\frac{\pi}{3}-\frac{5\pi}{6}+\pi n\\x = (-1)^{n}*\frac{\pi}{3}*\frac{1}{2}-\frac{5\pi}{6}*\frac{1}{2} +\pi n*\frac{1}{2}\\x = (-1)^{n}*\frac{\pi}{6}-\frac{5\pi}{12}+\frac{\pi n}{2}$n∈Z

Пусть n = 2, n = 3, n = 4, n = 5

x₁ = (-1)² × π/6 - 5π/12 + π×2/2 = 1 × π/6 - 5π/12 + π = π/6 - 5π/12 + π = 2π/12 - 5π/12 + 12π/12 = 9π/12 = 3π/4

x₂ = (-1)³ × π/6 - 5π/12 + π×3/2 = -1 × π/6 - 5π/12 + 3π/2 = -π/6 - 5π/12 + 3π/2 = -2π/12 - 5π/12 + 18π/12 = 11π/12

x₃ = (-1)⁴ × π/6 - 5π/12 + π×4/2 = 1 × π/6 - 5π/12 + 4π/2 = π/6 - 5π/12 + 4π/2 = 2π/12 - 5π/12 + 24π/12 = 21π/12 = 7π/4

x₄ = (-1)⁵ × π/6 - 5π/12 + π×5/2 = -1 × π/6 - 5π/12 + 5π/2 = -π/6 - 5π/12 + 5π/2 = -2π/12 - 5π/12 + 30π/12 = 23π/12

Ответ: x₁ = 3π/4, x₂ = 11π/12, x₃ = 7π/4, x₄ = 23π/12