прошу посодействовать с решением

3^(1/2 + cos x/3) + 6^(1/2) = 9^(1/2 + sin x/3)

3cos x/3 + 6 = 3sin x/3

cos x/3 gt; 0

sin x/3 gt; 0

3sin x/3 - 3cos x/3 = 6

-/2 + 2n lt; x/3 lt; /2 + 2n, n Z

2k lt; x/3 lt; + 2k, k Z

3/2 sin x/3 - 1/2 cos x/3 = 2/2

-1/2 + 2n lt; x/3 lt; 1/2 + 2n, n Z

2k lt; x/3 lt; 1 + 2k, k Z

sin x/3 cos /6 - cos x/3 sin /6 = 2/2

-3/2 + 6n lt; x lt; 3/2 + 6n, n Z

6k lt; x lt; 3 + 6k, k Z

sin(x/3 - /6) = 2/2

6k lt; x lt; 3/2 + 6k, k Z

x/3 - /6 = (-1)/4 + n, n Z

6k lt; x lt; 3/2 + 6k, k Z

x/3 - 1/6 = (-1)1/4 + n, n Z

6k lt; x lt; 3/2 + 6k, k Z

x = (-1)3/4 + 1/2 + 3n, n Z

6k lt; x lt; 3/2 + 6k, k Z

x = 3/4 + 1/2 + 6n, n Z

x = 5/4 + 6n, n Z