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arcsin(1/(x-2))>= 0

Solution

arcsin (\frac{1}{x - 2}) \geq 0

Solution

x \geq 3

Interval Notation

[3, \infty)

Solution steps

arcsin (\frac{1}{x - 2}) \geq 0

arcsin (x) \geq a

then

x \geq sin (a)

\frac{1}{x - 2} \geq sin (0)

sin (0) = 0

sin (0)

Use the following trivial identity:

sin (0) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

\frac{1}{x - 2} \geq 0

\frac{1}{x - 2} \geq 0 : x > 2

\frac{1}{x - 2} \geq 0

\frac{1}{a} \geq 0

then

a > 0

x - 2 > 0

Move

2

to the right side

x - 2 > 0

Add

2

to both sides

x - 2 + 2 > 0 + 2

Simplify

x > 2

x > 2

x > 2

Domain of

arcsin (\frac{1}{x - 2}) : x \leq 1 or x \geq 3

Domain definition

Find known functions domain restrictions:

x \leq 1 or x \geq 3

arcsin (f (x)) \Rightarrow - 1 \leq f (x) \leq 1

Solve

- 1 \leq \frac{1}{x - 2} \leq 1 : x \leq 1 or x \geq 3

- 1 \leq \frac{1}{x - 2} \leq 1

a \leq u \leq b

then

a \leq u and u \leq b

- 1 \leq \frac{1}{x - 2} and \frac{1}{x - 2} \leq 1

- 1 \leq \frac{1}{x - 2} : x \leq 1 or x > 2

- 1 \leq \frac{1}{x - 2}

Switch sides

\frac{1}{x - 2} \geq - 1

Rewrite in standard form

\frac{1}{x - 2} \geq - 1

Add

1

to both sides

\frac{1}{x - 2} + 1 \geq - 1 + 1

Simplify

\frac{1}{x - 2} + 1 \geq 0

Simplify

\frac{1}{x - 2} + 1 : \frac{x - 1}{x - 2}

\frac{1}{x - 2} + 1

Convert element to fraction:

1 = \frac{1 ( x - 2 )}{x - 2}

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

1 + 1 \cdot (x - 2) = x - 1

1 + 1 \cdot (x - 2)

1 \cdot (x - 2) = x - 2

1 \cdot (x - 2)

Multiply:

1 \cdot (x - 2) = (x - 2)

= (x - 2)

Remove parentheses:

(a) = a

= x - 2

= 1 + x - 2

Group like terms

= x + 1 - 2

Add/Subtract the numbers:

1 - 2 = - 1

= x - 1

= \frac{x - 1}{x - 2}

\frac{x - 1}{x - 2} \geq 0

\frac{x - 1}{x - 2} \geq 0

Identify the intervals

Find the signs of the factors of

\frac{x - 1}{x - 2}

Find the signs of

x - 1

x - 1 = 0 : x = 1

x - 1 = 0

Move

1

to the right side

x - 1 = 0

Add

1

to both sides

x - 1 + 1 = 0 + 1

Simplify

x = 1

x = 1

x - 1 < 0 : x < 1

x - 1 < 0

Move

1

to the right side

x - 1 < 0

Add

1

to both sides

x - 1 + 1 < 0 + 1

Simplify

x < 1

x < 1

x - 1 > 0 : x > 1

x - 1 > 0

Move

1

to the right side

x - 1 > 0

Add

1

to both sides

x - 1 + 1 > 0 + 1

Simplify

x > 1

x > 1

Find the signs of

x - 2

x - 2 = 0 : x = 2

x - 2 = 0

Move

2

to the right side

x - 2 = 0

Add

2

to both sides

x - 2 + 2 = 0 + 2

Simplify

x = 2

x = 2

x - 2 < 0 : x < 2

x - 2 < 0

Move

2

to the right side

x - 2 < 0

Add

2

to both sides

x - 2 + 2 < 0 + 2

Simplify

x < 2

x < 2

x - 2 > 0 : x > 2

x - 2 > 0

Move

2

to the right side

x - 2 > 0

Add

2

to both sides

x - 2 + 2 > 0 + 2

Simplify

x > 2

x > 2

Find singularity points

Find the zeros of the denominator

x - 2 : x = 2

x - 2 = 0

Move

2

to the right side

x - 2 = 0

Add

2

to both sides

x - 2 + 2 = 0 + 2

Simplify

x = 2

x = 2

Summarize in a table:

x - 1 x - 2 \frac{x - 1}{x - 2} x < 1 - - + x = 1 0 - 0 1 < x < 2 + - - x = 2 + 0 Undefined x > 2 + + +

Identify the intervals that satisfy the required condition:

\geq 0

x < 1 or x = 1 or x > 2

Merge Overlapping Intervals

x \leq 1 or x > 2

The union of two intervals is the set of numbers which are in either interval

x < 1

x = 1

x \leq 1

The union of two intervals is the set of numbers which are in either interval

x \leq 1

x > 2

x \leq 1 or x > 2

x \leq 1 or x > 2

x \leq 1 or x > 2

\frac{1}{x - 2} \leq 1 : x < 2 or x \geq 3

\frac{1}{x - 2} \leq 1

Rewrite in standard form

\frac{1}{x - 2} \leq 1

Subtract

1

from both sides

\frac{1}{x - 2} - 1 \leq 1 - 1

Simplify

\frac{1}{x - 2} - 1 \leq 0

Simplify

\frac{1}{x - 2} - 1 : \frac{- x + 3}{x - 2}

\frac{1}{x - 2} - 1

Convert element to fraction:

1 = \frac{1 ( x - 2 )}{x - 2}

= \frac{1}{x - 2} - \frac{1 \cdot ( x - 2 )}{x - 2}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 - 1 \cdot ( x - 2 )}{x - 2}

Multiply:

1 \cdot (x - 2) = (x - 2)

= \frac{1 - ( x - 2 )}{x - 2}

Expand

1 - (x - 2) : - x + 3

1 - (x - 2)

- (x - 2) : - x + 2

- (x - 2)

Distribute parentheses

= - (x) - (- 2)

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - x + 2

= 1 - x + 2

Simplify

1 - x + 2 : - x + 3

1 - x + 2

Group like terms

= - x + 1 + 2

Add the numbers:

1 + 2 = 3

= - x + 3

= - x + 3

= \frac{- x + 3}{x - 2}

\frac{- x + 3}{x - 2} \leq 0

\frac{- x + 3}{x - 2} \leq 0

Identify the intervals

Find the signs of the factors of

\frac{- x + 3}{x - 2}

Find the signs of

- x + 3

- x + 3 = 0 : x = 3

- x + 3 = 0

Move

3

to the right side

- x + 3 = 0

Subtract

3

from both sides

- x + 3 - 3 = 0 - 3

Simplify

- x = - 3

- x = - 3

Divide both sides by

- 1

- x = - 3

Divide both sides by

- 1

\frac{- x}{- 1} = \frac{- 3}{- 1}

Simplify

x = 3

x = 3

- x + 3 < 0 : x > 3

- x + 3 < 0

Move

3

to the right side

- x + 3 < 0

Subtract

3

from both sides

- x + 3 - 3 < 0 - 3

Simplify

- x < - 3

- x < - 3

Multiply both sides by

- 1

- x < - 3

Multiply both sides by -1 (reverse the inequality)

(- x) (- 1) > (- 3) (- 1)

Simplify

x > 3

x > 3

- x + 3 > 0 : x < 3

- x + 3 > 0

Move

3

to the right side

- x + 3 > 0

Subtract

3

from both sides

- x + 3 - 3 > 0 - 3

Simplify

- x > - 3

- x > - 3

Multiply both sides by

- 1

- x > - 3

Multiply both sides by -1 (reverse the inequality)

(- x) (- 1) < (- 3) (- 1)

Simplify

x < 3

x < 3

Find the signs of

x - 2

x - 2 = 0 : x = 2

x - 2 = 0

Move

2

to the right side

x - 2 = 0

Add

2

to both sides

x - 2 + 2 = 0 + 2

Simplify

x = 2

x = 2

x - 2 < 0 : x < 2

x - 2 < 0

Move

2

to the right side

x - 2 < 0

Add

2

to both sides

x - 2 + 2 < 0 + 2

Simplify

x < 2

x < 2

x - 2 > 0 : x > 2

x - 2 > 0

Move

2

to the right side

x - 2 > 0

Add

2

to both sides

x - 2 + 2 > 0 + 2

Simplify

x > 2

x > 2

Find singularity points

Find the zeros of the denominator

x - 2 : x = 2

x - 2 = 0

Move

2

to the right side

x - 2 = 0

Add

2

to both sides

x - 2 + 2 = 0 + 2

Simplify

x = 2

x = 2

Summarize in a table:

- x + 3 x - 2 \frac{- x + 3}{x - 2} x < 2 + - - x = 2 + 0 Undefined 2 < x < 3 + + + x = 3 0 + 0 x > 3 - + -

Identify the intervals that satisfy the required condition:

\leq 0

x < 2 or x = 3 or x > 3

Merge Overlapping Intervals

x < 2 or x = 3 or x > 3

The union of two intervals is the set of numbers which are in either interval

x < 2

x = 3

x < 2 or x = 3

The union of two intervals is the set of numbers which are in either interval

x < 2 or x = 3

x > 3

x < 2 or x \geq 3

x < 2 or x \geq 3

x < 2 or x \geq 3

Combine the intervals

(x \leq 1 or x > 2) and (x < 2 or x \geq 3)

Merge Overlapping Intervals

x \leq 1 or x > 2 and x < 2 or x \geq 3

The intersection of two intervals is the set of numbers which are in both intervals

x \leq 1 or x > 2

and

x < 2 or x \geq 3

x \leq 1 or x \geq 3

x \leq 1 or x \geq 3

Find undefined (singularity) points:

x = 2

arcsin (\frac{1}{x - 2})

Take the denominator(s) of

arcsin (\frac{1}{x - 2})

and compare to zero

Solve

x - 2 = 0 : x = 2

x - 2 = 0

Move

2

to the right side

x - 2 = 0

Add

2

to both sides

x - 2 + 2 = 0 + 2

Simplify

x = 2

x = 2

The following points are undefined

x = 2

Combine real regions and undefined points for final function domain

x \leq 1 or x \geq 3

Combine the intervals

x > 2 and x \leq 1 or x \geq 3

Merge Overlapping Intervals

x > 2 and x \leq 1 or x \geq 3

The intersection of two intervals is the set of numbers which are in both intervals

x > 2

and

x \leq 1 or x \geq 3

x \geq 3

x \geq 3

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