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Popular Trigonometry >

-2<= 2/(cos(x))<= 1

Solution

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

Solution

x = π + 2 πn

Decimal

x = 3.14159 \dots + 2 πn

Solution steps

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

a \leq u \leq b

then

a \leq u and u \leq b

- 2 \leq \frac{2}{cos ( x )} and \frac{2}{cos ( x )} \leq 1

- 2 \leq \frac{2}{cos ( x )} : - \frac{π}{2} + 2 πn < x < \frac{π}{2} + 2 πn or x = π + 2 πn

- 2 \leq \frac{2}{cos ( x )}

Switch sides

\frac{2}{cos ( x )} \geq - 2

Rewrite in standard form

\frac{2}{cos ( x )} \geq - 2

Add

2

to both sides

\frac{2}{cos ( x )} + 2 \geq - 2 + 2

Simplify

\frac{2}{cos ( x )} + 2 \geq 0

Simplify

\frac{2}{cos ( x )} + 2 : \frac{2 + 2 cos ( x )}{cos ( x )}

\frac{2}{cos ( x )} + 2

Convert element to fraction:

2 = \frac{2 cos ( x )}{cos ( x )}

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

Since the denominators are equal, combine the fractions:

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

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

\frac{2 + 2 cos ( x )}{cos ( x )} \geq 0

\frac{2 + 2 cos ( x )}{cos ( x )} \geq 0

Factor

\frac{2 + 2 cos ( x )}{cos ( x )} : \frac{2 ( cos ( x ) + 1 )}{cos ( x )}

\frac{2 + 2 cos ( x )}{cos ( x )}

Factor

2 cos (x) + 2 : 2 (cos (x) + 1)

2 cos (x) + 2

Factor out common term

2

= 2 (cos (x) + 1)

= \frac{2 ( cos ( x ) + 1 )}{cos ( x )}

\frac{2 ( cos ( x ) + 1 )}{cos ( x )} \geq 0

Divide both sides by

2

\frac{\frac{2 ( c o s ( x ) + 1 )}{c o s ( x )}}{2} \geq \frac{0}{2}

Simplify

\frac{cos ( x ) + 1}{cos ( x )} \geq 0

Identify the intervals

Find the signs of the factors of

\frac{cos ( x ) + 1}{cos ( x )}

Find the signs of

cos (x) + 1

cos (x) + 1 = 0 : cos (x) = - 1

cos (x) + 1 = 0

Move

1

to the right side

cos (x) + 1 = 0

Subtract

1

from both sides

cos (x) + 1 - 1 = 0 - 1

Simplify

cos (x) = - 1

cos (x) = - 1

cos (x) + 1 < 0 : cos (x) < - 1

cos (x) + 1 < 0

Move

1

to the right side

cos (x) + 1 < 0

Subtract

1

from both sides

cos (x) + 1 - 1 < 0 - 1

Simplify

cos (x) < - 1

cos (x) < - 1

cos (x) + 1 > 0 : cos (x) > - 1

cos (x) + 1 > 0

Move

1

to the right side

cos (x) + 1 > 0

Subtract

1

from both sides

cos (x) + 1 - 1 > 0 - 1

Simplify

cos (x) > - 1

cos (x) > - 1

Find the signs of

cos (x)

cos (x) = 0

cos (x) < 0

cos (x) > 0

Find singularity points

Find the zeros of the denominator

cos (x) : cos (x) = 0

Summarize in a table:

cos (x) + 1 cos (x) \frac{c o s ( x ) + 1}{c o s ( x )} cos (x) < - 1 - - + cos (x) = - 1 0 - 0 - 1 < cos (x) < 0 + - - cos (x) = 0 + 0 Undefined cos (x) > 0 + + +

Identify the intervals that satisfy the required condition:

\geq 0

cos (x) < - 1 or cos (x) = - 1 or cos (x) > 0

Merge Overlapping Intervals

cos (x) \leq - 1 or cos (x) > 0

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

cos (x) < - 1

cos (x) = - 1

cos (x) \leq - 1

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

cos (x) \leq - 1

cos (x) > 0

cos (x) \leq - 1 or cos (x) > 0

cos (x) \leq - 1 or cos (x) > 0

cos (x) \leq - 1 or cos (x) > 0

cos (x) \leq - 1 : x = π + 2 πn

cos (x) \leq - 1

For

cos (x) \leq a

, if

- 1 < a < 1

then

arccos (a) + 2 πn \leq x \leq 2 π - arccos (a) + 2 πn

arccos (- 1) + 2 πn \leq x \leq 2 π - arccos (- 1) + 2 πn

Simplify

arccos (- 1) : π

arccos (- 1)

Use the following trivial identity:

arccos (- 1) = π

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= π

Simplify

2 π - arccos (- 1) : π

2 π - arccos (- 1)

Use the following trivial identity:

arccos (- 1) = π

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 2 π - π

Add similar elements:

2 π - π = π

= π

π + 2 πn \leq x \leq π + 2 πn

Simplify

x = π + 2 πn

cos (x) > 0 : - \frac{π}{2} + 2 πn < x < \frac{π}{2} + 2 πn

cos (x) > 0

For

cos (x) > a

, if

- 1 \leq a < 1

then

- arccos (a) + 2 πn < x < arccos (a) + 2 πn

- arccos (0) + 2 πn < x < arccos (0) + 2 πn

Simplify

- arccos (0) : - \frac{π}{2}

- arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - \frac{π}{2}

Simplify

arccos (0) : \frac{π}{2}

arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{2}

- \frac{π}{2} + 2 πn < x < \frac{π}{2} + 2 πn

Combine the intervals

x = π + 2 πn or - \frac{π}{2} + 2 πn < x < \frac{π}{2} + 2 πn

Merge Overlapping Intervals

- \frac{π}{2} + 2 πn < x < \frac{π}{2} + 2 πn or x = π + 2 πn

\frac{2}{cos ( x )} \leq 1 : \frac{π}{2} + 2 πn < x < \frac{3 π}{2} + 2 πn

\frac{2}{cos ( x )} \leq 1

Rewrite in standard form

\frac{2}{cos ( x )} \leq 1

Subtract

1

from both sides

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

Simplify

\frac{2}{cos ( x )} - 1 \leq 0

Simplify

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

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

Convert element to fraction:

1 = \frac{1 cos ( x )}{cos ( x )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (x) = cos (x)

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

\frac{2 - cos ( x )}{cos ( x )} \leq 0

\frac{2 - cos ( x )}{cos ( x )} \leq 0

Identify the intervals

Find the signs of the factors of

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

Find the signs of

2 - cos (x)

2 - cos (x) = 0 : cos (x) = 2

2 - cos (x) = 0

Move

2

to the right side

2 - cos (x) = 0

Subtract

2

from both sides

2 - cos (x) - 2 = 0 - 2

Simplify

- cos (x) = - 2

- cos (x) = - 2

Divide both sides by

- 1

- cos (x) = - 2

Divide both sides by

- 1

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

Simplify

cos (x) = 2

cos (x) = 2

2 - cos (x) < 0 : cos (x) > 2

2 - cos (x) < 0

Move

2

to the right side

2 - cos (x) < 0

Subtract

2

from both sides

2 - cos (x) - 2 < 0 - 2

Simplify

- cos (x) < - 2

- cos (x) < - 2

Multiply both sides by

- 1

- cos (x) < - 2

Multiply both sides by -1 (reverse the inequality)

(- cos (x)) (- 1) > (- 2) (- 1)

Simplify

cos (x) > 2

cos (x) > 2

2 - cos (x) > 0 : cos (x) < 2

2 - cos (x) > 0

Move

2

to the right side

2 - cos (x) > 0

Subtract

2

from both sides

2 - cos (x) - 2 > 0 - 2

Simplify

- cos (x) > - 2

- cos (x) > - 2

Multiply both sides by

- 1

- cos (x) > - 2

Multiply both sides by -1 (reverse the inequality)

(- cos (x)) (- 1) < (- 2) (- 1)

Simplify

cos (x) < 2

cos (x) < 2

Find the signs of

cos (x)

cos (x) = 0

cos (x) < 0

cos (x) > 0

Find singularity points

Find the zeros of the denominator

cos (x) : cos (x) = 0

Summarize in a table:

2 - cos (x) cos (x) \frac{2 - c o s ( x )}{c o s ( x )} cos (x) < 0 + - - cos (x) = 0 + 0 Undefined 0 < cos (x) < 2 + + + cos (x) = 2 0 + 0 cos (x) > 2 - + -

Identify the intervals that satisfy the required condition:

\leq 0

cos (x) < 0 or cos (x) = 2 or cos (x) > 2

Merge Overlapping Intervals

cos (x) < 0 or cos (x) = 2 or cos (x) > 2

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

cos (x) < 0

cos (x) = 2

cos (x) < 0 or cos (x) = 2

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

cos (x) < 0 or cos (x) = 2

cos (x) > 2

cos (x) < 0 or cos (x) \geq 2

cos (x) < 0 or cos (x) \geq 2

cos (x) < 0 or cos (x) \geq 2

cos (x) < 0 : \frac{π}{2} + 2 πn < x < \frac{3 π}{2} + 2 πn

cos (x) < 0

For

cos (x) < a

, if

- 1 < a \leq 1

then

arccos (a) + 2 πn < x < 2 π - arccos (a) + 2 πn

arccos (0) + 2 πn < x < 2 π - arccos (0) + 2 πn

Simplify

arccos (0) : \frac{π}{2}

arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{2}

Simplify

2 π - arccos (0) : \frac{3 π}{2}

2 π - arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 2 π - \frac{π}{2}

Simplify

2 π - \frac{π}{2}

Convert element to fraction:

2 π = \frac{2 π 2}{2}

= \frac{2 π 2}{2} - \frac{π}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{2 π 2 - π}{2}

2 π 2 - π = 3 π

2 π 2 - π

Multiply the numbers:

2 \cdot 2 = 4

= 4 π - π

Add similar elements:

4 π - π = 3 π

= 3 π

= \frac{3 π}{2}

= \frac{3 π}{2}

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

cos (x) \geq 2 :

False for all

x \in R

cos (x) \geq 2

Range of

cos (x) : - 1 \leq cos (x) \leq 1

Function range definition

The range of the basic

cos

function is

- 1 \leq cos (x) \leq 1

- 1 \leq cos (x) \leq 1

cos (x) \geq 2 and - 1 \leq cos (x) \leq 1 :

False

Let

y = cos (x)

Combine the intervals

y \geq 2 and - 1 \leq y \leq 1

Merge Overlapping Intervals

y \geq 2 and - 1 \leq y \leq 1

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

y \geq 2

and

- 1 \leq y \leq 1

False for all y \in R

False for all y \in R

No Solution for x \in R

False for all x \in R

Combine the intervals

\frac{π}{2} + 2 πn < x < \frac{3 π}{2} + 2 πn or False for all x \in R

Merge Overlapping Intervals

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

Combine the intervals

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

Merge Overlapping Intervals

x = π + 2 πn

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