What is the domain of f(x)=(2sin(x)-1)/(sin(x)+1) ?

The domain of f(x)=(2sin(x)-1)/(sin(x)+1) is 2pin<= x<(3pi)/2+2pin\lor (3pi)/2+2pin<x<2pi+2pin

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domain of f(x)=(2sin(x)-1)/(sin(x)+1)

domain

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

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

Interval Notation

[2 πn, \frac{3 π}{2} + 2 πn) \cup (\frac{3 π}{2} + 2 πn, 2 π + 2 πn)

Solution steps

Take the denominator(s) of

\frac{2 sin ( x ) - 1}{sin ( x ) + 1}

and compare to zero:

x = \frac{3 π}{2} + 2 πn

Periodicity of

\frac{2 sin ( x ) - 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{3 π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < x < 2 π + 2 πn

f (x) = 1 - 2 lo g_{2} (- x)

f (x) = x^{2} + 6 x + 10

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

7 x + 6

7 x + 8

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