Question

Решите пожалуйста нужна помощь

Answer 1

Решение:

$cosx=\frac{\sqrt{2}}{2}\\x = +-arccos(\frac{\sqrt{2}}{2}) + 2\pi n, n \in Z \\x = +-\frac{\pi}{4}+2\pi n, n \in Z$

Пусть n = 0,и n = 1:

x₁ = -π/4 + 2×0×π = -π/4

x₂ = π/4 + 2×0×π = π/4

x₃ = -π/4 + 2×1×π = -π/4 + 2π = 7π/4

Ответ: x₁ = -π/4; x₂ = π/4; x₃ = 7π/4