5. Найдите наибольший отрицательный коре cos(x+π/12) = √2/2.

cos(x + π/12) = √2 / 2

x + π/12 = ± arccos(√2 / 2) + 2πn, n \in Z

x + π/12 = ± π/4 + 2πn, n \in Z

1) x + π/12 = π/4 + 2πn, n \in Z

x = π/4 - π/12 + 2πn, n \in Z

x = (3π - π) / 12 + 2πn, n \in Z

x = 2π/12 + 2πn, n \in Z

x = π/6 + 2πn, n \in Z

2) x + π/12 = -π/4 + 2πn, n \in Z

x = -π/4 - π/12 + 2πn, n \in Z

x = (-3π - π) / 12 + 2πn, n \in Z

x = -4π/12 + 2πn, n \in Z

x = -π/3 + 2πn, n \in Z

Ищем наибольший отрицательный корень:

1) x = π/6 + 2πn, n \in Z

n = -1: x = π/6 - 2π = (π - 12π) / 6 = -11π/6

n = 0: x = π/6

2) x = -π/3 + 2πn, n \in Z

n = 0: x = -π/3

n = -1: x = -π/3 - 2π = (-π - 6π) / 3 = -7π/3

-π/3 > -11π/6 и -π/3 > -7π/3

Ответ: $$-\frac{π}{3}$$