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

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

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

domain

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

2 πn \leq x < \frac{π}{4} + 2 πn or \frac{π}{4} + 2 πn < x < \frac{3 π}{4} + 2 πn or \frac{3 π}{4} + 2 πn < x < \frac{5 π}{4} + 2 πn or \frac{5 π}{4} + 2 πn < x < \frac{7 π}{4} + 2 πn or \frac{7 π}{4} + 2 πn < x < 2 π + 2 πn

Interval Notation

[2 πn, \frac{π}{4} + 2 πn) \cup (\frac{π}{4} + 2 πn, \frac{3 π}{4} + 2 πn) \cup (\frac{3 π}{4} + 2 πn, \frac{5 π}{4} + 2 πn) \cup (\frac{5 π}{4} + 2 πn, \frac{7 π}{4} + 2 πn) \cup (\frac{7 π}{4} + 2 πn, 2 π + 2 πn)

Solution steps

Take the denominator(s) of

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

and compare to zero:

x = \frac{π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Periodicity of

\frac{1 - sin ( x )}{cos ( 2 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}{4} + 2 πn or \frac{π}{4} + 2 πn < x < \frac{3 π}{4} + 2 πn or \frac{3 π}{4} + 2 πn < x < \frac{5 π}{4} + 2 πn or \frac{5 π}{4} + 2 πn < x < \frac{7 π}{4} + 2 πn or \frac{7 π}{4} + 2 πn < x < 2 π + 2 πn

f (x) = - x^{\frac{1}{2}} + 5

f (x) = \frac{1 - 5 x}{4 + x}

y = 6 x + 3

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