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Popular Trigonometry >

5-5cos^2(x)+2cos(x)>= 0

Solution

5 - 5 cos^{2} (x) + 2 cos (x) \geq 0

Solution

- arccos (\frac{- 26 + 1}{5}) + 2 πn \leq x \leq arccos (\frac{- 26 + 1}{5}) + 2 πn

Interval Notation

[- arccos (\frac{- 26 + 1}{5}) + 2 πn, arccos (\frac{- 26 + 1}{5}) + 2 πn]

Decimal

- 2.53186 \dots + 2 πn \leq x \leq 2.53186 \dots + 2 πn

Solution steps

5 - 5 cos^{2} (x) + 2 cos (x) \geq 0

Let:

u = cos (x)

5 - 5 u^{2} + 2 u \geq 0

5 - 5 u^{2} + 2 u \geq 0 : \frac{- 26 + 1}{5} \leq u \leq \frac{26 + 1}{5}

5 - 5 u^{2} + 2 u \geq 0

Complete the square

5 - 5 u^{2} + 2 u : - 5 (u - \frac{1}{5})^{2} + \frac{26}{5}

5 - 5 u^{2} + 2 u

Write in the standard form

a x^{2} + b x + c

- 5 u^{2} + 2 u + 5

Write

- 5 u^{2} + 2 u + 5

in the form:

x^{2} + 2 a x + a^{2}

Factor out

- 5

- 5 (u^{2} - \frac{2 u}{5} - 1)

2 a = - \frac{2}{5} : a = - \frac{1}{5}

2 a = - \frac{2}{5}

Divide both sides by

2

2 a = - \frac{2}{5}

Divide both sides by

2

\frac{2 a}{2} = \frac{- \frac{2}{5}}{2}

Simplify

\frac{2 a}{2} = \frac{- \frac{2}{5}}{2}

Simplify

\frac{2 a}{2} : a

\frac{2 a}{2}

Divide the numbers:

\frac{2}{2} = 1

= a

Simplify

\frac{- \frac{2}{5}}{2} : - \frac{1}{5}

\frac{- \frac{2}{5}}{2}

Apply the fraction rule:

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

= - \frac{\frac{2}{5}}{2}

Apply the fraction rule:

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

\frac{\frac{2}{5}}{2} = \frac{2}{5 \cdot 2}

= - \frac{2}{5 \cdot 2}

Multiply the numbers:

5 \cdot 2 = 10

= - \frac{2}{10}

Cancel the common factor:

2

= - \frac{1}{5}

a = - \frac{1}{5}

a = - \frac{1}{5}

a = - \frac{1}{5}

Add and subtract

(- \frac{1}{5})^{2}

- 5 (u^{2} - \frac{2 u}{5} - 1 + (- \frac{1}{5})^{2} - (- \frac{1}{5})^{2})

x^{2} + 2 a x + a^{2} = (x + a)^{2}

u^{2} - \frac{2}{5} u + (- \frac{1}{5})^{2} = (u - \frac{1}{5})^{2}

- 5 ((u - \frac{1}{5})^{2} - 1 - (- \frac{1}{5})^{2})

Simplify

- 5 (u - \frac{1}{5})^{2} + \frac{26}{5}

- 5 (u - \frac{1}{5})^{2} + \frac{26}{5} \geq 0

Move

\frac{26}{5}

to the right side

- 5 (u - \frac{1}{5})^{2} + \frac{26}{5} \geq 0

Subtract

\frac{26}{5}

from both sides

- 5 (u - \frac{1}{5})^{2} + \frac{26}{5} - \frac{26}{5} \geq 0 - \frac{26}{5}

Simplify

- 5 (u - \frac{1}{5})^{2} \geq - \frac{26}{5}

- 5 (u - \frac{1}{5})^{2} \geq - \frac{26}{5}

Multiply both sides by

- 1

- 5 (u - \frac{1}{5})^{2} \geq - \frac{26}{5}

Multiply both sides by -1 (reverse the inequality)

(- 5 (u - \frac{1}{5})^{2}) (- 1) \leq (- \frac{26}{5}) (- 1)

Simplify

5 (u - \frac{1}{5})^{2} \leq \frac{26}{5}

5 (u - \frac{1}{5})^{2} \leq \frac{26}{5}

Divide both sides by

5

5 (u - \frac{1}{5})^{2} \leq \frac{26}{5}

Divide both sides by

5

\frac{5 ( u - \frac{1}{5} ) ^{2}}{5} \leq \frac{\frac{26}{5}}{5}

Simplify

\frac{5 ( u - \frac{1}{5} ) ^{2}}{5} \leq \frac{\frac{26}{5}}{5}

Simplify

\frac{5 ( u - \frac{1}{5} ) ^{2}}{5} : (u - \frac{1}{5})^{2}

\frac{5 ( u - \frac{1}{5} ) ^{2}}{5}

Divide the numbers:

\frac{5}{5} = 1

= (u - \frac{1}{5})^{2}

Simplify

\frac{\frac{26}{5}}{5} : \frac{26}{25}

\frac{\frac{26}{5}}{5}

Apply the fraction rule:

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

= \frac{26}{5 \cdot 5}

Multiply the numbers:

5 \cdot 5 = 25

= \frac{26}{25}

(u - \frac{1}{5})^{2} \leq \frac{26}{25}

(u - \frac{1}{5})^{2} \leq \frac{26}{25}

(u - \frac{1}{5})^{2} \leq \frac{26}{25}

For

u^{n} \leq a

, if

n

is even then

- \frac{26}{25} \leq u - \frac{1}{5} \leq \frac{26}{25}

a \leq u \leq b

then

a \leq u and u \leq b

- \frac{26}{25} \leq u - \frac{1}{5} and u - \frac{1}{5} \leq \frac{26}{25}

- \frac{26}{25} \leq u - \frac{1}{5} : u \geq \frac{- 26 + 1}{5}

- \frac{26}{25} \leq u - \frac{1}{5}

Switch sides

u - \frac{1}{5} \geq - \frac{26}{25}

Simplify

\frac{26}{25} : \frac{26}{5}

\frac{26}{25}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{26}{25}

25 = 5

25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

= \frac{26}{5}

u - \frac{1}{5} \geq - \frac{26}{5}

Move

\frac{1}{5}

to the right side

u - \frac{1}{5} \geq - \frac{26}{5}

Add

\frac{1}{5}

to both sides

u - \frac{1}{5} + \frac{1}{5} \geq - \frac{26}{5} + \frac{1}{5}

Simplify

u - \frac{1}{5} + \frac{1}{5} \geq - \frac{26}{5} + \frac{1}{5}

Simplify

u - \frac{1}{5} + \frac{1}{5} : u

u - \frac{1}{5} + \frac{1}{5}

Add similar elements:

- \frac{1}{5} + \frac{1}{5} \geq 0

= u

Simplify

- \frac{26}{5} + \frac{1}{5} : \frac{- 26 + 1}{5}

- \frac{26}{5} + \frac{1}{5}

Apply rule

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

= \frac{- 26 + 1}{5}

u \geq \frac{- 26 + 1}{5}

u \geq \frac{- 26 + 1}{5}

u \geq \frac{- 26 + 1}{5}

u - \frac{1}{5} \leq \frac{26}{25} : u \leq \frac{26 + 1}{5}

u - \frac{1}{5} \leq \frac{26}{25}

Apply radical rule:

assuming

a \geq 0, b \geq 0

u - \frac{1}{5} \leq \frac{26}{25}

25 = 5

25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u - \frac{1}{5} \leq \frac{26}{5}

Move

\frac{1}{5}

to the right side

u - \frac{1}{5} \leq \frac{26}{5}

Add

\frac{1}{5}

to both sides

u - \frac{1}{5} + \frac{1}{5} \leq \frac{26}{5} + \frac{1}{5}

Simplify

u - \frac{1}{5} + \frac{1}{5} \leq \frac{26}{5} + \frac{1}{5}

Simplify

u - \frac{1}{5} + \frac{1}{5} : u

u - \frac{1}{5} + \frac{1}{5}

Add similar elements:

- \frac{1}{5} + \frac{1}{5} \leq 0

= u

Simplify

\frac{26}{5} + \frac{1}{5} : \frac{26 + 1}{5}

\frac{26}{5} + \frac{1}{5}

Apply rule

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

= \frac{26 + 1}{5}

u \leq \frac{26 + 1}{5}

u \leq \frac{26 + 1}{5}

u \leq \frac{26 + 1}{5}

Combine the intervals

u \geq \frac{- 26 + 1}{5} and u \leq \frac{26 + 1}{5}

Merge Overlapping Intervals

u \geq \frac{- 26 + 1}{5} and u \leq \frac{26 + 1}{5}

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

u \geq \frac{- 26 + 1}{5}

and

u \leq \frac{26 + 1}{5}

\frac{- 26 + 1}{5} \leq u \leq \frac{26 + 1}{5}

\frac{- 26 + 1}{5} \leq u \leq \frac{26 + 1}{5}

\frac{- 26 + 1}{5} \leq u \leq \frac{26 + 1}{5}

Substitute back

u = cos (x)

\frac{- 26 + 1}{5} \leq cos (x) \leq \frac{26 + 1}{5}

a \leq u \leq b

then

a \leq u and u \leq b

\frac{- 26 + 1}{5} \leq cos (x) and cos (x) \leq \frac{26 + 1}{5}

\frac{- 26 + 1}{5} \leq cos (x) : - arccos (\frac{- 26 + 1}{5}) + 2 πn \leq x \leq arccos (\frac{- 26 + 1}{5}) + 2 πn

\frac{- 26 + 1}{5} \leq cos (x)

Switch sides

cos (x) \geq \frac{- 26 + 1}{5}

For

cos (x) \geq a

, if

- 1 < a < 1

then

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

- arccos (\frac{- 26 + 1}{5}) + 2 πn \leq x \leq arccos (\frac{- 26 + 1}{5}) + 2 πn

cos (x) \leq \frac{26 + 1}{5} :

True for all

x \in R

cos (x) \leq \frac{26 + 1}{5}

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) \leq \frac{26 + 1}{5} and - 1 \leq cos (x) \leq 1 : - 1 \leq cos (x) \leq 1

Let

y = cos (x)

Combine the intervals

y \leq \frac{26 + 1}{5} and - 1 \leq y \leq 1

Merge Overlapping Intervals

y \leq \frac{26 + 1}{5} and - 1 \leq y \leq 1

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

y \leq \frac{26 + 1}{5}

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

Combine the intervals

- arccos (\frac{- 26 + 1}{5}) + 2 πn \leq x \leq arccos (\frac{- 26 + 1}{5}) + 2 πn and True for all x \in R

Merge Overlapping Intervals

- arccos (\frac{- 26 + 1}{5}) + 2 πn \leq x \leq arccos (\frac{- 26 + 1}{5}) + 2 πn

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