What is the domain of f(x)= 1/(sin(x)) ?

The domain of f(x)= 1/(sin(x)) is 2pin<x<pi+2pin\lor pi+2pin<x<2pi+2pin

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domain of f(x)= 1/(sin(x))

domain

f (x) = \frac{1}{sin ( x )}

2 πn < x < π + 2 πn or π + 2 πn < x < 2 π + 2 πn

Interval Notation

(2 πn, π + 2 πn) \cup (π + 2 πn, 2 π + 2 πn)

Solution steps

Take the denominator(s) of

\frac{1}{sin ( x )}

and compare to zero:

x = 2 πn, x = π + 2 πn

Periodicity of

\frac{1}{sin ( x )} : 2 π

Combine period:

2 πn \leq x < 2 π + 2 πn

and undefined points.

The domain is

2 πn < x < π + 2 πn or π + 2 πn < x < 2 π + 2 πn

x^{2} + 5

f (x) = 9 + x^{2}

arcsin (\frac{1}{1 - x})

\frac{t - 2}{4 t - 12}

f (x) = \frac{5 x + 2}{x - 5}

What is the domain of f(x)= 1/(sin(x)) ?
The domain of f(x)= 1/(sin(x)) is 2pin<x<pi+2pin\lor pi+2pin<x<2pi+2pin