To solve the first equation, let's substitute y = ctg(x/3):

y^2 - 6y + 5 = 0

This is a quadratic equation that can be factored as:

(y - 5)(y - 1) = 0

Thus, y = 5 or y = 1.

Now, substitute y back in terms of ctg(x/3):

ctg(x/3) = 5
x/3 = arctan(1/5)
x = 3 * arctan(1/5)

Or,

ctg(x/3) = 1
x/3 = π/4
x = 3π/4

For the second equation, we substitute z = ctg(x + π/6):

z^2 - 6z + 5 = 0

(z - 5)(z - 1) = 0

Thus, z = 5 or z = 1.

Substitute z back in terms of ctg(x + π/6):

ctg(x + π/6) = 5
x + π/6 = arctan(5)
x = arctan(5) - π/6

Or,

ctg(x + π/6) = 1
x + π/6 = π/4
x = π/4 - π/6

Therefore, the solutions are x = 3 * arctan(1/5), 3π/4, arctan(5) - π/6, and π/4 - π/6.