Sin(x + 3π/2) = -cos(x)
Сделаем замену:
9^(cosx) = t
t + 1/t = 10/3
3t² + 3 = 10t
3t² - 10t + 3 = 0
D = 100 - 36 = 64
t1 = (10 - 8)/6 = 1/3
t2 = (10 + 8)/6 = 3
вернемся к замене:
1) 9^(cosx) = 1/3
cosx = -1/2
x = ±2π/3 + 2πk, k∈Z
2) 9^(cosx) = 3
cosx = 1/2
x = ±π/3 + 2πk, k∈Z
Ответ: ±2π/3 + 2πk, k∈Z, ±π/3 + 2πk, k∈Z