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

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

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

domain

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

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}{1 - sin ( x )}

and compare to zero:

x = \frac{π}{2} + 2 πn

Periodicity of

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

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

- x^{2} + 12 x - 34

f (x) = \frac{x}{3 x ^{2} + 2 x - 8}

\frac{x ^{3}}{x ^{4} - 1}

f (x) = 2 x + 8, - 3 < x < 2

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