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

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

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

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

Period : 2 π

Domain : 2 πn < x < 2 π + 2 πn

Range : - 2 \leq f (x) \leq 2

X Intercepts : (π + 2 πn, 0)

Asymptotes : None

Extreme Points : Minimum (\frac{π}{2} + 2 πn, - 2), Maximum (\frac{3 π}{2} + 2 πn, 2)

Interval Notation

Domain : (2 πn, 2 π + 2 πn)

Range : [- 2, 2]

Solution steps

Periodicity of

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

Domain of

\frac{sin ( 2 x ) - 2 sin ( x )}{1 - cos ( x )} : 2 πn < x < 2 π + 2 πn

Range of

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

Axis interception points of

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

X Intercepts

: (π + 2 πn, 0)

Asymptotes of

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

None

Extreme Points of

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

Minimum

(\frac{π}{2} + 2 πn, - 2),

Maximum

(\frac{3 π}{2} + 2 πn, 2)

f (x) = - x^{2} - 6 x + 3

f (x) = - ∣ x - 3 ∣

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

f (x) = - 1.5 sin (\frac{2 x}{7} - \frac{π}{7}) - 3

f (s) = s^{2} - s + 1

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