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Popular >

4cos(x/3+pi/4)+sqrt(12)>= 0

Solution

4 cos (\frac{x}{3} + \frac{π}{4}) + 12 \geq 0

Solution

- \frac{13 π}{4} + 6 πn \leq x \leq \frac{7 π}{4} + 6 πn

Interval Notation

[- \frac{13 π}{4} + 6 πn, \frac{7 π}{4} + 6 πn]

Decimal Notation

- 10.21017 \dots + 6 πn \leq x \leq 5.49778 \dots + 6 πn

Solution steps

4 cos (\frac{x}{3} + \frac{π}{4}) + 12 \geq 0

12 = 23

12

Prime factorization of

12 : 2^{2} \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

= 3 2^{2}

Apply radical rule:

2^{2} = 2

= 23

4 cos (\frac{x}{3} + \frac{π}{4}) + 23 \geq 0

Move

23

to the right side

4 cos (\frac{x}{3} + \frac{π}{4}) + 23 \geq 0

Subtract

23

from both sides

4 cos (\frac{x}{3} + \frac{π}{4}) + 23 - 23 \geq 0 - 23

Simplify

4 cos (\frac{x}{3} + \frac{π}{4}) \geq - 23

4 cos (\frac{x}{3} + \frac{π}{4}) \geq - 23

Divide both sides by

4

4 cos (\frac{x}{3} + \frac{π}{4}) \geq - 23

Divide both sides by

4

\frac{4 cos ( \frac{x}{3} + \frac{π}{4} )}{4} \geq \frac{- 2 3}{4}

Simplify

\frac{4 cos ( \frac{x}{3} + \frac{π}{4} )}{4} \geq \frac{- 2 3}{4}

Simplify

\frac{4 cos ( \frac{x}{3} + \frac{π}{4} )}{4} : cos (\frac{x}{3} + \frac{π}{4})

\frac{4 cos ( \frac{x}{3} + \frac{π}{4} )}{4}

Divide the numbers:

\frac{4}{4} = 1

= cos (\frac{x}{3} + \frac{π}{4})

Simplify

\frac{- 2 3}{4} : - \frac{3}{2}

\frac{- 2 3}{4}

Apply the fraction rule:

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

= - \frac{2 3}{4}

Cancel the common factor:

2

= - \frac{3}{2}

cos (\frac{x}{3} + \frac{π}{4}) \geq - \frac{3}{2}

cos (\frac{x}{3} + \frac{π}{4}) \geq - \frac{3}{2}

cos (\frac{x}{3} + \frac{π}{4}) \geq - \frac{3}{2}

For

cos (x) \geq a

, if

- 1 < a < 1

then

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

- arccos (- \frac{3}{2}) + 2 πn \leq (\frac{x}{3} + \frac{π}{4}) \leq arccos (- \frac{3}{2}) + 2 πn

a \leq u \leq b

then

a \leq u and u \leq b

- arccos (- \frac{3}{2}) + 2 πn \leq \frac{x}{3} + \frac{π}{4} and \frac{x}{3} + \frac{π}{4} \leq arccos (- \frac{3}{2}) + 2 πn

- arccos (- \frac{3}{2}) + 2 πn \leq \frac{x}{3} + \frac{π}{4} : x \geq 6 πn - \frac{13 π}{4}

- arccos (- \frac{3}{2}) + 2 πn \leq \frac{x}{3} + \frac{π}{4}

Switch sides

\frac{x}{3} + \frac{π}{4} \geq - arccos (- \frac{3}{2}) + 2 πn

Simplify

- arccos (- \frac{3}{2}) + 2 πn : - \frac{5 π}{6} + 2 πn

- arccos (- \frac{3}{2}) + 2 πn

Use the following trivial identity:

arccos (- \frac{3}{2}) = \frac{5 π}{6}

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{5 π}{6} + 2 πn

\frac{x}{3} + \frac{π}{4} \geq - \frac{5 π}{6} + 2 πn

Move

\frac{π}{4}

to the right side

\frac{x}{3} + \frac{π}{4} \geq - \frac{5 π}{6} + 2 πn

Subtract

\frac{π}{4}

from both sides

\frac{x}{3} + \frac{π}{4} - \frac{π}{4} \geq - \frac{5 π}{6} + 2 πn - \frac{π}{4}

Simplify

\frac{x}{3} + \frac{π}{4} - \frac{π}{4} \geq - \frac{5 π}{6} + 2 πn - \frac{π}{4}

Simplify

\frac{x}{3} + \frac{π}{4} - \frac{π}{4} : \frac{x}{3}

\frac{x}{3} + \frac{π}{4} - \frac{π}{4}

Add similar elements:

\frac{π}{4} - \frac{π}{4} \geq 0

= \frac{x}{3}

Simplify

- \frac{5 π}{6} + 2 πn - \frac{π}{4} : 2 πn - \frac{13 π}{12}

- \frac{5 π}{6} + 2 πn - \frac{π}{4}

Group like terms

= 2 πn - \frac{π}{4} - \frac{5 π}{6}

Least Common Multiplier of

4, 6 : 12

4, 6

Least Common Multiplier (LCM)

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Prime factorization of

6 : 2 \cdot 3

6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3

Multiply each factor the greatest number of times it occurs in either

4

6

= 2 \cdot 2 \cdot 3

Multiply the numbers:

2 \cdot 2 \cdot 3 = 12

= 12

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

12

For

\frac{π}{4} :

multiply the denominator and numerator by

3

\frac{π}{4} = \frac{π 3}{4 \cdot 3} = \frac{π 3}{12}

For

\frac{5 π}{6} :

multiply the denominator and numerator by

2

\frac{5 π}{6} = \frac{5 π 2}{6 \cdot 2} = \frac{10 π}{12}

= - \frac{π 3}{12} - \frac{10 π}{12}

Since the denominators are equal, combine the fractions:

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

= \frac{- π 3 - 10 π}{12}

Add similar elements:

- 3 π - 10 π = - 13 π

= \frac{- 13 π}{12}

Apply the fraction rule:

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

= 2 πn - \frac{13 π}{12}

\frac{x}{3} \geq 2 πn - \frac{13 π}{12}

\frac{x}{3} \geq 2 πn - \frac{13 π}{12}

\frac{x}{3} \geq 2 πn - \frac{13 π}{12}

Multiply both sides by

3

\frac{x}{3} \geq 2 πn - \frac{13 π}{12}

Multiply both sides by

3

\frac{3 x}{3} \geq 3 \cdot 2 πn - 3 \cdot \frac{13 π}{12}

Simplify

\frac{3 x}{3} \geq 3 \cdot 2 πn - 3 \cdot \frac{13 π}{12}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

3 \cdot 2 πn - 3 \cdot \frac{13 π}{12} : 6 πn - \frac{13 π}{4}

3 \cdot 2 πn - 3 \cdot \frac{13 π}{12}

3 \cdot 2 πn = 6 πn

3 \cdot 2 πn

Multiply the numbers:

3 \cdot 2 = 6

= 6 πn

3 \cdot \frac{13 π}{12} = \frac{13 π}{4}

3 \cdot \frac{13 π}{12}

Multiply fractions:

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

= \frac{13 π 3}{12}

Multiply the numbers:

13 \cdot 3 = 39

= \frac{39 π}{12}

Cancel the common factor:

3

= \frac{13 π}{4}

= 6 πn - \frac{13 π}{4}

x \geq 6 πn - \frac{13 π}{4}

x \geq 6 πn - \frac{13 π}{4}

x \geq 6 πn - \frac{13 π}{4}

\frac{x}{3} + \frac{π}{4} \leq arccos (- \frac{3}{2}) + 2 πn : x \leq 6 πn + \frac{7 π}{4}

\frac{x}{3} + \frac{π}{4} \leq arccos (- \frac{3}{2}) + 2 πn

Simplify

arccos (- \frac{3}{2}) + 2 πn : \frac{5 π}{6} + 2 πn

arccos (- \frac{3}{2}) + 2 πn

Use the following trivial identity:

arccos (- \frac{3}{2}) = \frac{5 π}{6}

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{5 π}{6} + 2 πn

\frac{x}{3} + \frac{π}{4} \leq \frac{5 π}{6} + 2 πn

Move

\frac{π}{4}

to the right side

\frac{x}{3} + \frac{π}{4} \leq \frac{5 π}{6} + 2 πn

Subtract

\frac{π}{4}

from both sides

\frac{x}{3} + \frac{π}{4} - \frac{π}{4} \leq \frac{5 π}{6} + 2 πn - \frac{π}{4}

Simplify

\frac{x}{3} + \frac{π}{4} - \frac{π}{4} \leq \frac{5 π}{6} + 2 πn - \frac{π}{4}

Simplify

\frac{x}{3} + \frac{π}{4} - \frac{π}{4} : \frac{x}{3}

\frac{x}{3} + \frac{π}{4} - \frac{π}{4}

Add similar elements:

\frac{π}{4} - \frac{π}{4} \leq 0

= \frac{x}{3}

Simplify

\frac{5 π}{6} + 2 πn - \frac{π}{4} : 2 πn + \frac{7 π}{12}

\frac{5 π}{6} + 2 πn - \frac{π}{4}

Group like terms

= 2 πn - \frac{π}{4} + \frac{5 π}{6}

Least Common Multiplier of

4, 6 : 12

4, 6

Least Common Multiplier (LCM)

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Prime factorization of

6 : 2 \cdot 3

6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3

Multiply each factor the greatest number of times it occurs in either

4

6

= 2 \cdot 2 \cdot 3

Multiply the numbers:

2 \cdot 2 \cdot 3 = 12

= 12

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

12

For

\frac{π}{4} :

multiply the denominator and numerator by

3

\frac{π}{4} = \frac{π 3}{4 \cdot 3} = \frac{π 3}{12}

For

\frac{5 π}{6} :

multiply the denominator and numerator by

2

\frac{5 π}{6} = \frac{5 π 2}{6 \cdot 2} = \frac{10 π}{12}

= - \frac{π 3}{12} + \frac{10 π}{12}

Since the denominators are equal, combine the fractions:

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

= \frac{- π 3 + 10 π}{12}

Add similar elements:

- 3 π + 10 π = 7 π

= 2 πn + \frac{7 π}{12}

\frac{x}{3} \leq 2 πn + \frac{7 π}{12}

\frac{x}{3} \leq 2 πn + \frac{7 π}{12}

\frac{x}{3} \leq 2 πn + \frac{7 π}{12}

Multiply both sides by

3

\frac{x}{3} \leq 2 πn + \frac{7 π}{12}

Multiply both sides by

3

\frac{3 x}{3} \leq 3 \cdot 2 πn + 3 \cdot \frac{7 π}{12}

Simplify

\frac{3 x}{3} \leq 3 \cdot 2 πn + 3 \cdot \frac{7 π}{12}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

3 \cdot 2 πn + 3 \cdot \frac{7 π}{12} : 6 πn + \frac{7 π}{4}

3 \cdot 2 πn + 3 \cdot \frac{7 π}{12}

3 \cdot 2 πn = 6 πn

3 \cdot 2 πn

Multiply the numbers:

3 \cdot 2 = 6

= 6 πn

3 \cdot \frac{7 π}{12} = \frac{7 π}{4}

3 \cdot \frac{7 π}{12}

Multiply fractions:

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

= \frac{7 π 3}{12}

Multiply the numbers:

7 \cdot 3 = 21

= \frac{21 π}{12}

Cancel the common factor:

3

= \frac{7 π}{4}

= 6 πn + \frac{7 π}{4}

x \leq 6 πn + \frac{7 π}{4}

x \leq 6 πn + \frac{7 π}{4}

x \leq 6 πn + \frac{7 π}{4}

Combine the intervals

x \geq 6 πn - \frac{13 π}{4} and x \leq 6 πn + \frac{7 π}{4}

Merge Overlapping Intervals

- \frac{13 π}{4} + 6 πn \leq x \leq \frac{7 π}{4} + 6 πn

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