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2-10cos((pit)/(12))>= 0

Solution

2 - 10 cos (\frac{π t}{12}) \geq 0

Solution

\frac{12 arccos ( \frac{1}{5} )}{π} + 24 n \leq t \leq \frac{24 π - 12 arccos ( \frac{1}{5} )}{π} + 24 n

Interval Notation

[\frac{12 arccos ( \frac{1}{5} )}{π} + 24 n, \frac{24 π - 12 arccos ( \frac{1}{5} )}{π} + 24 n]

Decimal Notation

5.23086 \dots + 24 n \leq t \leq 18.76913 \dots + 24 n

Solution steps

2 - 10 cos (\frac{π t}{12}) \geq 0

Move

2

to the right side

2 - 10 cos (\frac{π t}{12}) \geq 0

Subtract

2

from both sides

2 - 10 cos (\frac{π t}{12}) - 2 \geq 0 - 2

Simplify

- 10 cos (\frac{π t}{12}) \geq - 2

- 10 cos (\frac{π t}{12}) \geq - 2

Multiply both sides by

- 1

- 10 cos (\frac{π t}{12}) \geq - 2

Multiply both sides by -1 (reverse the inequality)

(- 10 cos (\frac{π t}{12})) (- 1) \leq (- 2) (- 1)

Simplify

10 cos (\frac{π t}{12}) \leq 2

10 cos (\frac{π t}{12}) \leq 2

Divide both sides by

10

10 cos (\frac{π t}{12}) \leq 2

Divide both sides by

10

\frac{10 cos ( \frac{π t}{12} )}{10} \leq \frac{2}{10}

Simplify

cos (\frac{π t}{12}) \leq \frac{1}{5}

cos (\frac{π t}{12}) \leq \frac{1}{5}

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{1}{5}) + 2 πn \leq \frac{π t}{12} \leq 2 π - arccos (\frac{1}{5}) + 2 πn

a \leq u \leq b

then

a \leq u and u \leq b

arccos (\frac{1}{5}) + 2 πn \leq \frac{π t}{12} and \frac{π t}{12} \leq 2 π - arccos (\frac{1}{5}) + 2 πn

arccos (\frac{1}{5}) + 2 πn \leq \frac{π t}{12} : t \geq \frac{12 arccos ( \frac{1}{5} )}{π} + 24 n

arccos (\frac{1}{5}) + 2 πn \leq \frac{π t}{12}

Switch sides

\frac{π t}{12} \geq arccos (\frac{1}{5}) + 2 πn

Multiply both sides by

12

\frac{π t}{12} \geq arccos (\frac{1}{5}) + 2 πn

Multiply both sides by

12

\frac{12 π t}{12} \geq 12 arccos (\frac{1}{5}) + 12 \cdot 2 πn

Simplify

\frac{12 π t}{12} \geq 12 arccos (\frac{1}{5}) + 12 \cdot 2 πn

Simplify

\frac{12 π t}{12} : π t

\frac{12 π t}{12}

Divide the numbers:

\frac{12}{12} = 1

= π t

Simplify

12 arccos (\frac{1}{5}) + 12 \cdot 2 πn : 12 arccos (\frac{1}{5}) + 24 πn

12 arccos (\frac{1}{5}) + 12 \cdot 2 πn

Multiply the numbers:

12 \cdot 2 = 24

= 12 arccos (\frac{1}{5}) + 24 πn

π t \geq 12 arccos (\frac{1}{5}) + 24 πn

π t \geq 12 arccos (\frac{1}{5}) + 24 πn

π t \geq 12 arccos (\frac{1}{5}) + 24 πn

Divide both sides by

π

π t \geq 12 arccos (\frac{1}{5}) + 24 πn

Divide both sides by

π

\frac{π t}{π} \geq \frac{12 arccos ( \frac{1}{5} )}{π} + \frac{24 πn}{π}

Simplify

t \geq \frac{12 arccos ( \frac{1}{5} )}{π} + 24 n

t \geq \frac{12 arccos ( \frac{1}{5} )}{π} + 24 n

\frac{π t}{12} \leq 2 π - arccos (\frac{1}{5}) + 2 πn : t \leq \frac{24 π - 12 arccos ( \frac{1}{5} )}{π} + 24 n

\frac{π t}{12} \leq 2 π - arccos (\frac{1}{5}) + 2 πn

Multiply both sides by

12

\frac{π t}{12} \leq 2 π - arccos (\frac{1}{5}) + 2 πn

Multiply both sides by

12

\frac{12 π t}{12} \leq 12 \cdot 2 π - 12 arccos (\frac{1}{5}) + 12 \cdot 2 πn

Simplify

\frac{12 π t}{12} \leq 12 \cdot 2 π - 12 arccos (\frac{1}{5}) + 12 \cdot 2 πn

Simplify

\frac{12 π t}{12} : π t

\frac{12 π t}{12}

Divide the numbers:

\frac{12}{12} = 1

= π t

Simplify

12 \cdot 2 π - 12 arccos (\frac{1}{5}) + 12 \cdot 2 πn : 24 π - 12 arccos (\frac{1}{5}) + 24 πn

12 \cdot 2 π - 12 arccos (\frac{1}{5}) + 12 \cdot 2 πn

Multiply the numbers:

12 \cdot 2 = 24

= 24 π - 12 arccos (\frac{1}{5}) + 24 πn

π t \leq 24 π - 12 arccos (\frac{1}{5}) + 24 πn

π t \leq 24 π - 12 arccos (\frac{1}{5}) + 24 πn

π t \leq 24 π - 12 arccos (\frac{1}{5}) + 24 πn

Divide both sides by

π

π t \leq 24 π - 12 arccos (\frac{1}{5}) + 24 πn

Divide both sides by

π

\frac{π t}{π} \leq \frac{24 π}{π} - \frac{12 arccos ( \frac{1}{5} )}{π} + \frac{24 πn}{π}

Simplify

\frac{π t}{π} \leq \frac{24 π}{π} - \frac{12 arccos ( \frac{1}{5} )}{π} + \frac{24 πn}{π}

Simplify

\frac{π t}{π} : t

\frac{π t}{π}

Cancel the common factor:

π

= t

Simplify

\frac{24 π}{π} - \frac{12 arccos ( \frac{1}{5} )}{π} + \frac{24 πn}{π} : 24 - \frac{12 arccos ( \frac{1}{5} )}{π} + 24 n

\frac{24 π}{π} - \frac{12 arccos ( \frac{1}{5} )}{π} + \frac{24 πn}{π}

Cancel

\frac{24 π}{π} : 24

\frac{24 π}{π}

Cancel the common factor:

π

= 24

= 24 - \frac{12 arccos ( \frac{1}{5} )}{π} + \frac{24 πn}{π}

Cancel

\frac{24 πn}{π} : 24 n

\frac{24 πn}{π}

Cancel the common factor:

π

= 24 n

= 24 - \frac{12 arccos ( \frac{1}{5} )}{π} + 24 n

t \leq 24 - \frac{12 arccos ( \frac{1}{5} )}{π} + 24 n

t \leq 24 - \frac{12 arccos ( \frac{1}{5} )}{π} + 24 n

Simplify

24 - \frac{12 arccos ( \frac{1}{5} )}{π} : \frac{24 π - 12 arccos ( \frac{1}{5} )}{π}

24 - \frac{12 arccos ( \frac{1}{5} )}{π}

Convert element to fraction:

24 = \frac{24 π}{π}

= \frac{24 π}{π} - \frac{12 arccos ( \frac{1}{5} )}{π}

Since the denominators are equal, combine the fractions:

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

= \frac{24 π - 12 arccos ( \frac{1}{5} )}{π}

t \leq \frac{24 π - 12 arccos ( \frac{1}{5} )}{π} + 24 n

t \leq \frac{24 π - 12 arccos ( \frac{1}{5} )}{π} + 24 n

Combine the intervals

t \geq \frac{12 arccos ( \frac{1}{5} )}{π} + 24 n and t \leq \frac{24 π - 12 arccos ( \frac{1}{5} )}{π} + 24 n

Merge Overlapping Intervals

\frac{12 arccos ( \frac{1}{5} )}{π} + 24 n \leq t \leq \frac{24 π - 12 arccos ( \frac{1}{5} )}{π} + 24 n

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