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

The domain of 1/(1+sin(x)) is 2pin<= x<(3pi)/2+2pin\lor (3pi)/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{3 π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < x < 2 π + 2 πn

Interval Notation

Solution steps

Take the denominator(s) of

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

and compare to zero:

x = \frac{3 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{3 π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < x < 2 π + 2 πn

f (x) = 2 x^{2} - 6 x + 4

lo g_{4} (x - 3) + lo g_{4} (x + 1)

x (x - 2)

f (x) = 5 - 5 - x

\frac{x - 1}{x ^{2} + 5 x + 6}

What is the domain of 1/(1+sin(x)) ?
The domain of 1/(1+sin(x)) is 2pin<= x<(3pi)/2+2pin\lor (3pi)/2+2pin<x<2pi+2pin