English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

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

Interval Notation

(\frac{2 π}{3} + 2 πn, \frac{4 π}{3} + 2 πn)

Decimal Notation

2.09439 \dots + 2 πn < x < 4.18879 \dots + 2 πn

Solution steps

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

a \leq u < b

then

a \leq u and u < b

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

- 1 \leq cos (x) :

True for all

x \in R

- 1 \leq cos (x)

Switch sides

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

Let

y = cos (x)

Combine the intervals

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

Merge Overlapping Intervals

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

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

y \geq - 1

and

- 1 \leq y \leq 1

- 1 \leq y \leq 1

- 1 \leq y \leq 1

True for all x

True for all x \in R

cos (x) < - \frac{1}{2} : \frac{2 π}{3} + 2 πn < x < \frac{4 π}{3} + 2 πn

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

For

cos (x) < a

, if

- 1 < a \leq 1

then

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

arccos (- \frac{1}{2}) + 2 πn < x < 2 π - arccos (- \frac{1}{2}) + 2 πn

Simplify

arccos (- \frac{1}{2}) : \frac{2 π}{3}

arccos (- \frac{1}{2})

Use the following trivial identity:

arccos (- \frac{1}{2}) = \frac{2 π}{3}

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{2 π}{3}

Simplify

2 π - arccos (- \frac{1}{2}) : \frac{4 π}{3}

2 π - arccos (- \frac{1}{2})

Use the following trivial identity:

arccos (- \frac{1}{2}) = \frac{2 π}{3}

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{2 π}{3}

Simplify

2 π - \frac{2 π}{3}

Convert element to fraction:

2 π = \frac{2 π 3}{3}

= \frac{2 π 3}{3} - \frac{2 π}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{2 π 3 - 2 π}{3}

2 π 3 - 2 π = 4 π

2 π 3 - 2 π

Multiply the numbers:

2 \cdot 3 = 6

= 6 π - 2 π

Add similar elements:

6 π - 2 π = 4 π

= 4 π

= \frac{4 π}{3}

= \frac{4 π}{3}

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

Combine the intervals

True for all x \in R and \frac{2 π}{3} + 2 πn < x < \frac{4 π}{3} + 2 πn

Merge Overlapping Intervals

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

Popular Examples

cos(θ)= 2/3 \land sin(θ)<0

cos (θ) = \frac{2}{3} and sin (θ) < 0

sec(360)<= θ<= 360

sec^{\circ} (36 0^{\circ}) \leq θ \leq 360

-1<(0.2)/(4*cos^2(x)-3)<0

- 1 < \frac{0.2}{4 \cdot cos ^{2} ( x ) - 3} < 0

-180<tan(x)<180

- 18 0^{\circ} < tan (x) < 18 0^{\circ}

-pi/2 <arctan(y)< pi/2

- \frac{π}{2} < arctan (y) < \frac{π}{2}