pre-calc 12 help on trig identities?

tan[x - (π/2)] = 1/√3
tan[x - (π/2)] = tan(π/6)
x - (π/2) = nπ + (π/6)
x = nπ + (π/6) + (π/2)
x = nπ + (2π/3)  where n ∊ Z

tan(x - π/2) = 1/√3

so, x - π/2 = π/6 + nπ....for n ∈ ℤ 

Hence, x = 2π/3 + nπ...general solution

Note: With n = 0 and n = 1 we have:

x = 2π/3 and 5π/3...for 0 < x < 2π

:)>

tan(t) = 1/sqrt(3) when t = pi/6 + pi * k, where k is an integer.  That's from the unit circle.

tan(x - pi/2) = 1/sqrt(3)
tan(x - pi/2) = tan(pi/6 + pi * k)
x - pi/2 = pi/6 + pi * k
x = pi/2 + pi/6 + pi * k
x = 3pi/6 + pi/6 + pi * k
x = 4pi/6 + pi * k
x = 2pi/3 + pi * k
x = (pi/3) * (2 + 3k)

Again, k is an integer