What is the domain of f(x)= 1/(tan(2x)-(sqrt(3))/3) ?

The domain of f(x)= 1/(tan(2x)-(sqrt(3))/3) is

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domain of f(x)= 1/(tan(2x)-(sqrt(3))/3)

domain

f (x) = \frac{1}{tan ( 2 x ) - \frac{3}{3}}

\frac{π}{2} n \leq x < \frac{π}{12} + \frac{π}{2} n or \frac{π}{12} + \frac{π}{2} n < x < \frac{π}{4} + \frac{π}{2} n or \frac{π}{4} + \frac{π}{2} n < x < \frac{π}{2} + \frac{π}{2} n

Interval Notation

[\frac{π}{2} n, \frac{π}{12} + \frac{π}{2} n) \cup (\frac{π}{12} + \frac{π}{2} n, \frac{π}{4} + \frac{π}{2} n) \cup (\frac{π}{4} + \frac{π}{2} n, \frac{π}{2} + \frac{π}{2} n)

Solution steps

Take the denominator(s) of

\frac{1}{tan ( 2 x ) - \frac{3}{3}}

and compare to zero:

x = \frac{π}{12} + \frac{π}{2} n, x = \frac{π}{4} + \frac{π}{2} n

Periodicity of

\frac{1}{tan ( 2 x ) - \frac{3}{3}} : \frac{π}{2}

Combine period:

\frac{π}{2} n \leq x < \frac{π}{2} + \frac{π}{2} n

and undefined points.

The domain is

\frac{π}{2} n \leq x < \frac{π}{12} + \frac{π}{2} n or \frac{π}{12} + \frac{π}{2} n < x < \frac{π}{4} + \frac{π}{2} n or \frac{π}{4} + \frac{π}{2} n < x < \frac{π}{2} + \frac{π}{2} n

(x + 4)^{4}

2 + 2, x > 2

f (x) = ln (\frac{x - 2}{x + 3})

tan (x + 2)

f (x) = 8 - 3 x - (x - 4)

What is the domain of f(x)= 1/(tan(2x)-(sqrt(3))/3) ?
The domain of f(x)= 1/(tan(2x)-(sqrt(3))/3) is