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Popular Trigonometry >

sec^2(x)<1

Solution

sec^{2} (x) < 1

Solution

False for all x \in R

Solution steps

sec^{2} (x) < 1

Express with sin, cos

sec^{2} (x) < 1

Use the basic trigonometric identity:

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

(\frac{1}{cos ( x )})^{2} < 1

(\frac{1}{cos ( x )})^{2} < 1

For

u^{n} < a

, if

n

is even then

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

a < u < b

then

a < u and u < b

- 1 < \frac{1}{cos ( x )} and \frac{1}{cos ( x )} < 1

- 1 < \frac{1}{cos ( x )} : cos (x) < - 1 or cos (x) > 0

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

Switch sides

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

Rewrite in standard form

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

Add

1

to both sides

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

Simplify

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

Simplify

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

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

Convert element to fraction:

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

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

Multiply:

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

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

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

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

Identify the intervals

Find the signs of the factors of

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

Find the signs of

1 + cos (x)

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

1 + cos (x) = 0

Move

1

to the right side

1 + cos (x) = 0

Subtract

1

from both sides

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

Simplify

cos (x) = - 1

cos (x) = - 1

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

1 + cos (x) < 0

Move

1

to the right side

1 + cos (x) < 0

Subtract

1

from both sides

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

Simplify

cos (x) < - 1

cos (x) < - 1

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

1 + cos (x) > 0

Move

1

to the right side

1 + cos (x) > 0

Subtract

1

from both sides

1 + cos (x) - 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:

1 + cos (x) cos (x) \frac{1 + c o s ( x )}{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:

> 0

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

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

\frac{1}{cos ( x )} < 1 : cos (x) < 0 or cos (x) > 1

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

Rewrite in standard form

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

Subtract

1

from both sides

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

Simplify

\frac{1}{cos ( x )} - 1 < 0

Simplify

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

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

Convert element to fraction:

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

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

Multiply:

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

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

\frac{1 - cos ( x )}{cos ( x )} < 0

\frac{1 - cos ( x )}{cos ( x )} < 0

Identify the intervals

Find the signs of the factors of

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

Find the signs of

1 - cos (x)

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

1 - cos (x) = 0

Move

1

to the right side

1 - cos (x) = 0

Subtract

1

from both sides

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

Simplify

- cos (x) = - 1

- cos (x) = - 1

Divide both sides by

- 1

- cos (x) = - 1

Divide both sides by

- 1

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

Simplify

cos (x) = 1

cos (x) = 1

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

1 - cos (x) < 0

Move

1

to the right side

1 - cos (x) < 0

Subtract

1

from both sides

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

Simplify

- cos (x) < - 1

- cos (x) < - 1

Multiply both sides by

- 1

- cos (x) < - 1

Multiply both sides by -1 (reverse the inequality)

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

Simplify

cos (x) > 1

cos (x) > 1

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

1 - cos (x) > 0

Move

1

to the right side

1 - cos (x) > 0

Subtract

1

from both sides

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

Simplify

- cos (x) > - 1

- cos (x) > - 1

Multiply both sides by

- 1

- cos (x) > - 1

Multiply both sides by -1 (reverse the inequality)

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

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

Identify the intervals that satisfy the required condition:

< 0

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

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

Combine the intervals

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

Merge Overlapping Intervals

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

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

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

and

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

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

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

cos (x) < - 1 :

False for all

x \in R

cos (x) < - 1

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

False

Let

y = cos (x)

Combine the intervals

y < - 1 and - 1 \leq y \leq 1

Merge Overlapping Intervals

y < - 1 and - 1 \leq y \leq 1

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

y < - 1

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

cos (x) > 1 :

False for all

x \in R

cos (x) > 1

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

False

Let

y = cos (x)

Combine the intervals

y > 1 and - 1 \leq y \leq 1

Merge Overlapping Intervals

y > 1 and - 1 \leq y \leq 1

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

y > 1

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

False for all x \in R or False for all x \in R

Merge Overlapping Intervals

False for all x \in R or False for all x \in R

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

False for all

x \in R

False for all

x \in R

False for all x \in R

No Solution for x \in R

False for all x \in R

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