100 ! 9^(cosx)+9^(sin(x+3π/2))=10/3

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