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

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

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

domain

y = \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

1.5 x^{2} - x + 3.7

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

f (x) = \frac{( x ^{2} - 9 )}{x + 4}

y = - 5 - 3 x - 15 - 6

f (x) = x^{2} + 10 x + 2

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