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

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

domain

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

and compare to zero:

x = \frac{π}{2} + 2 πn

Periodicity of

\frac{2 cos ( 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{π}{2} + 2 πn or \frac{π}{2} + 2 πn < x < 2 π + 2 πn

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

f (x) = 8 + 7 x - 8

2 x^{2} + 3 x - 20

f (x) = x \cdot e^{x}

tan (x - \frac{π}{2})

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