Я так поняла, что нужно решить и отобрать корни, принадлежащие указанному промежутку
sin3x = √2/2
3x = (-1)ⁿarcsin(√2/2) + πn, n ∈ Z
3x = (-1)ⁿ · π/4 + πn, n ∈ Z
x = (-1)ⁿ · π/12 + πn/3, n ∈ Z
при n = 0 x = π/12 ∈ [-3π/2; 3π/2]
при n = 1 x = -π/12 + π/3 = 3π/12 = π/4 ∈ [-3π/2; 3π/2]
при n = 2 x = π/12 + 2π/3 = 9π/12 = 3π/4 ∈ [-3π/2; 3π/2]
при n = 3 x = -π/12 + π = 11π/12 ∈ [-3π/2; 3π/2]
при n = 4 x = π/12 + 4π/3 = 17π/12 ∈ [-3π/2; 3π/2]
при n = 5 x = -π/12 + 5π/3 = 19π/12 ∉ [-3π/2; 3π/2]
при n = -1 x = -π/12 - π/3 = -5π/12 ∈ [-3π/2; 3π/2]
при n = -2 x = π/12 - 2π/3 = -7π/12 ∈ [-3π/2; 3π/2]
при n = -3 x = -π/12 - π = -13π/12 ∈ [-3π/2; 3π/2]
при n = -4 x = π/12 - 4π/3 = -15π/12 = - 5π/4 ∈ [-3π/2; 3π/2]
при n = -5 x = -π/12 - 5π/3 = -21π/12 = -7π/4 ∉ [-3π/2; 3π/2]