1. а) arccos(√2/2) + arccos(-1/2) - arccos0 = π/4 + π - π/3 - π/2 = 5π/4 - 5π/6 = 15π/12 - 10π/12 = 5π/12
б) arccos(ctg(3π/4)) = arccos(-1) = π
в) sin(arccos(-√3/2) = sin(π - π/6) = sin(π/6) = 1/2
2. а) 2cost = √3
cost = √3/2
t = ±π/6 + 2πn, n ∈ Z
б) cost = -1
t = π + 2πn, n ∈ Z
в) cost = √17/4
Сравним √17/4 с 1.
Возведём в квадрат:
17/16 > 1. Значит, √17/4 > 1.
Тогда уравнение не имеет корней, т.к. cost ∈ [-1; 1]
г) cost = -√15/4
t = ±arccos(-√15/4) + 2πn, n ∈ Z