English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

2cos(x)+cos^2(x)>0

Solution

2 cos (x) + cos^{2} (x) > 0

Solution

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

Interval Notation

(- \frac{π}{2} + 2 πn, \frac{π}{2} + 2 πn)

Decimal

- 1.57079 \dots + 2 πn < x < 1.57079 \dots + 2 πn

Solution steps

2 cos (x) + cos^{2} (x) > 0

Let:

u = cos (x)

2 u + u^{2} > 0

2 u + u^{2} > 0 : u < - 2 or u > 0

2 u + u^{2} > 0

Factor

2 u + u^{2} : u (u + 2)

2 u + u^{2}

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

u^{2} = uu

= uu + 2 u

Factor out common term

u

= u (u + 2)

u (u + 2) > 0

Identify the intervals

Find the signs of the factors of

u (u + 2)

Find the signs of

u

u = 0

u < 0

u > 0

Find the signs of

u + 2

u + 2 = 0 : u = - 2

u + 2 = 0

Move

2

to the right side

u + 2 = 0

Subtract

2

from both sides

u + 2 - 2 = 0 - 2

Simplify

u = - 2

u = - 2

u + 2 < 0 : u < - 2

u + 2 < 0

Move

2

to the right side

u + 2 < 0

Subtract

2

from both sides

u + 2 - 2 < 0 - 2

Simplify

u < - 2

u < - 2

u + 2 > 0 : u > - 2

u + 2 > 0

Move

2

to the right side

u + 2 > 0

Subtract

2

from both sides

u + 2 - 2 > 0 - 2

Simplify

u > - 2

u > - 2

Summarize in a table:

u u + 2 u (u + 2) u < - 2 - - + u = - 2 - 00 - 2 < u < 0 - + - u = 0 0 + 0 u > 0 + + +

Identify the intervals that satisfy the required condition:

> 0

u < - 2 or u > 0

u < - 2 or u > 0

u < - 2 or u > 0

Substitute back

u = cos (x)

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

cos (x) < - 2 :

False for all

x \in R

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

False

Let

y = cos (x)

Combine the intervals

y < - 2 and - 1 \leq y \leq 1

Merge Overlapping Intervals

y < - 2 and - 1 \leq y \leq 1

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

y < - 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

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

False for all x \in R 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

Popular Examples

tan(x)>-2/3

sin(x)-cos(x)>= 1

3cos(t)>= 0

cos(x)-(sqrt(2))/2 <= 0

4*sin^2(X)-2<0