English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1/(cos(x))>= sqrt(2)

Solution

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

Solution

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

Interval Notation

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

Decimal Notation

0.78539 \dots + 2 πn \leq x < 1.57079 \dots + 2 πn or 4.71238 \dots + 2 πn < x \leq 5.49778 \dots + 2 πn

Solution steps

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

Rewrite in standard form

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

Subtract

2

from both sides

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

Simplify

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

Simplify

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

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

Convert element to fraction:

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

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

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

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

Identify the intervals

Find the signs of the factors of

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

Find the signs of

1 - 2 cos (x)

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

1 - 2 cos (x) = 0

Move

1

to the right side

1 - 2 cos (x) = 0

Subtract

1

from both sides

1 - 2 cos (x) - 1 = 0 - 1

Simplify

- 2 cos (x) = - 1

- 2 cos (x) = - 1

Divide both sides by

- 2

- 2 cos (x) = - 1

Divide both sides by

- 2

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

Simplify

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

Simplify

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

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

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

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

Cancel the common factor:

2

= cos (x)

Simplify

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

\frac{- 1}{- 2}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{1}{2}

Rationalize

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

= \frac{2}{2}

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

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

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

1 - 2 cos (x) < 0 : cos (x) > \frac{2}{2}

1 - 2 cos (x) < 0

Move

1

to the right side

1 - 2 cos (x) < 0

Subtract

1

from both sides

1 - 2 cos (x) - 1 < 0 - 1

Simplify

- 2 cos (x) < - 1

- 2 cos (x) < - 1

Multiply both sides by

- 1

- 2 cos (x) < - 1

Multiply both sides by -1 (reverse the inequality)

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

Simplify

2 cos (x) > 1

2 cos (x) > 1

Divide both sides by

2

2 cos (x) > 1

Divide both sides by

2

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

Simplify

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

Simplify

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

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

Cancel the common factor:

2

= cos (x)

Simplify

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

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

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

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

1 - 2 cos (x) > 0 : cos (x) < \frac{2}{2}

1 - 2 cos (x) > 0

Move

1

to the right side

1 - 2 cos (x) > 0

Subtract

1

from both sides

1 - 2 cos (x) - 1 > 0 - 1

Simplify

- 2 cos (x) > - 1

- 2 cos (x) > - 1

Multiply both sides by

- 1

- 2 cos (x) > - 1

Multiply both sides by -1 (reverse the inequality)

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

Simplify

2 cos (x) < 1

2 cos (x) < 1

Divide both sides by

2

2 cos (x) < 1

Divide both sides by

2

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

Simplify

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

Simplify

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

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

Cancel the common factor:

2

= cos (x)

Simplify

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

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

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

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

1 - 2 cos (x) cos (x) \frac{1 - 2 c o s ( x )}{c o s ( x )} cos (x) < 0 + - - cos (x) = 0 + 0 Undefined 0 < cos (x) < \frac{2}{2} + + + cos (x) = \frac{2}{2} 0 + 0 cos (x) > \frac{2}{2} - + -

Identify the intervals that satisfy the required condition:

\geq 0

0 < cos (x) < \frac{2}{2} or cos (x) = \frac{2}{2}

Merge Overlapping Intervals

0 < cos (x) < \frac{2}{2} or cos (x) = \frac{2}{2}

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

0 < cos (x) < \frac{2}{2}

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

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

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

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

a < u \leq b

then

a < u and u \leq b

0 < cos (x) and cos (x) \leq \frac{2}{2}

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

0 < cos (x)

Switch sides

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

cos (x) \leq \frac{2}{2} : \frac{π}{4} + 2 πn \leq x \leq \frac{7 π}{4} + 2 πn

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

For

cos (x) \leq a

, if

- 1 < a < 1

then

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

arccos (\frac{2}{2}) + 2 πn \leq x \leq 2 π - arccos (\frac{2}{2}) + 2 πn

Simplify

arccos (\frac{2}{2}) : \frac{π}{4}

arccos (\frac{2}{2})

Use the following trivial identity:

arccos (\frac{2}{2}) = \frac{π}{4}

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}{4}

Simplify

2 π - arccos (\frac{2}{2}) : \frac{7 π}{4}

2 π - arccos (\frac{2}{2})

Use the following trivial identity:

arccos (\frac{2}{2}) = \frac{π}{4}

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}{4}

Simplify

2 π - \frac{π}{4}

Convert element to fraction:

2 π = \frac{2 π 4}{4}

= \frac{2 π 4}{4} - \frac{π}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{2 π 4 - π}{4}

2 π 4 - π = 7 π

2 π 4 - π

Multiply the numbers:

2 \cdot 4 = 8

= 8 π - π

Add similar elements:

8 π - π = 7 π

= 7 π

= \frac{7 π}{4}

= \frac{7 π}{4}

\frac{π}{4} + 2 πn \leq x \leq \frac{7 π}{4} + 2 πn

Combine the intervals

- \frac{π}{2} + 2 πn < x < \frac{π}{2} + 2 πn and \frac{π}{4} + 2 πn \leq x \leq \frac{7 π}{4} + 2 πn

Merge Overlapping Intervals

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

Popular Examples

cos(2x)> 1/(sqrt(2))

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

sin(x)>= cos(x)

sin (x) \geq cos (x)

cos(x)<(sqrt(3))/2

cos (x) < \frac{3}{2}

2sin^2(x)-5sin(x)-3>= 0,0<= x<= 2pi

2 sin^{2} (x) - 5 sin (x) - 3 \geq 0, 0 \leq x \leq 2 π

cos(3x)>0

cos (3 x) > 0