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

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

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

domain

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

2 πn \leq x < \frac{π}{2} + 2 πn or \frac{π}{2} + 2 πn < x < 2 π + 2 πn

Interval Notation

[2 πn, \frac{π}{2} + 2 πn) \cup (\frac{π}{2} + 2 πn, 2 π + 2 πn)

Solution steps

Take the denominator(s) of

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

and compare to zero:

x = \frac{π}{2} + 2 πn

Periodicity of

\frac{1}{sin ( x ) - 1} : 2 π

Combine period:

2 πn \leq x < 2 π + 2 πn

and undefined points.

The domain is

2 πn \leq x < \frac{π}{2} + 2 πn or \frac{π}{2} + 2 πn < x < 2 π + 2 πn

13 - (13 - 2 x)

f (x) = \frac{x}{x ^{2} - 4}

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

f (x) = \frac{x ^{2} - 4}{x ^{2} + 1}

x^{2} - 6 x + 8

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