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Popular >

2cos(3x-1/2)>= sqrt(2)

Solution

2 cos (3 x - \frac{1}{2}) \geq 2

Solution

\frac{- π + 2}{12} + \frac{2 π}{3} n \leq x \leq \frac{π + 2}{12} + \frac{2 π}{3} n

Interval Notation

[\frac{- π + 2}{12} + \frac{2 π}{3} n, \frac{π + 2}{12} + \frac{2 π}{3} n]

Decimal Notation

- 0.09513 \dots + \frac{2 π}{3} n \leq x \leq 0.42846 \dots + \frac{2 π}{3} n

Solution steps

2 cos (3 x - \frac{1}{2}) \geq 2

Divide both sides by

2

2 cos (3 x - \frac{1}{2}) \geq 2

Divide both sides by

2

\frac{2 cos ( 3 x - \frac{1}{2} )}{2} \geq \frac{2}{2}

Simplify

cos (3 x - \frac{1}{2}) \geq \frac{2}{2}

cos (3 x - \frac{1}{2}) \geq \frac{2}{2}

For

cos (x) \geq a

, if

- 1 < a < 1

then

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

- arccos (\frac{2}{2}) + 2 πn \leq (3 x - \frac{1}{2}) \leq arccos (\frac{2}{2}) + 2 πn

a \leq u \leq b

then

a \leq u and u \leq b

- arccos (\frac{2}{2}) + 2 πn \leq 3 x - \frac{1}{2} and 3 x - \frac{1}{2} \leq arccos (\frac{2}{2}) + 2 πn

- arccos (\frac{2}{2}) + 2 πn \leq 3 x - \frac{1}{2} : x \geq \frac{- π + 2}{12} + \frac{2 π}{3} n

- arccos (\frac{2}{2}) + 2 πn \leq 3 x - \frac{1}{2}

Switch sides

3 x - \frac{1}{2} \geq - arccos (\frac{2}{2}) + 2 πn

Simplify

- arccos (\frac{2}{2}) + 2 πn : - \frac{π}{4} + 2 πn

- arccos (\frac{2}{2}) + 2 πn

Use the following trivial identity:

arccos (\frac{2}{2}) = \frac{π}{4}

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}{4} + 2 πn

3 x - \frac{1}{2} \geq - \frac{π}{4} + 2 πn

Move

\frac{1}{2}

to the right side

3 x - \frac{1}{2} \geq - \frac{π}{4} + 2 πn

Add

\frac{1}{2}

to both sides

3 x - \frac{1}{2} + \frac{1}{2} \geq - \frac{π}{4} + 2 πn + \frac{1}{2}

Simplify

3 x - \frac{1}{2} + \frac{1}{2} \geq - \frac{π}{4} + 2 πn + \frac{1}{2}

Simplify

3 x - \frac{1}{2} + \frac{1}{2} : 3 x

3 x - \frac{1}{2} + \frac{1}{2}

Add similar elements:

- \frac{1}{2} + \frac{1}{2} \geq 0

= 3 x

Simplify

- \frac{π}{4} + 2 πn + \frac{1}{2} : 2 πn - \frac{π}{4} + \frac{1}{2}

- \frac{π}{4} + 2 πn + \frac{1}{2}

Group like terms

= 2 πn - \frac{π}{4} + \frac{1}{2}

Could not simplify further

= 2 πn - \frac{π}{4} + \frac{1}{2}

3 x \geq 2 πn - \frac{π}{4} + \frac{1}{2}

3 x \geq 2 πn - \frac{π}{4} + \frac{1}{2}

3 x \geq 2 πn - \frac{π}{4} + \frac{1}{2}

Divide both sides by

3

3 x \geq 2 πn - \frac{π}{4} + \frac{1}{2}

Divide both sides by

3

\frac{3 x}{3} \geq \frac{2 πn}{3} - \frac{\frac{π}{4}}{3} + \frac{\frac{1}{2}}{3}

Simplify

\frac{3 x}{3} \geq \frac{2 πn}{3} - \frac{\frac{π}{4}}{3} + \frac{\frac{1}{2}}{3}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{2 πn}{3} - \frac{\frac{π}{4}}{3} + \frac{\frac{1}{2}}{3} : \frac{2 πn}{3} - \frac{π}{12} + \frac{1}{6}

\frac{2 πn}{3} - \frac{\frac{π}{4}}{3} + \frac{\frac{1}{2}}{3}

\frac{\frac{π}{4}}{3} = \frac{π}{12}

\frac{\frac{π}{4}}{3}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 3}

Multiply the numbers:

4 \cdot 3 = 12

= \frac{π}{12}

\frac{\frac{1}{2}}{3} = \frac{1}{6}

\frac{\frac{1}{2}}{3}

Apply the fraction rule:

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

= \frac{1}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{1}{6}

= \frac{2 πn}{3} - \frac{π}{12} + \frac{1}{6}

x \geq \frac{2 πn}{3} - \frac{π}{12} + \frac{1}{6}

x \geq \frac{2 πn}{3} - \frac{π}{12} + \frac{1}{6}

Simplify

- \frac{π}{12} + \frac{1}{6} : \frac{- π + 2}{12}

- \frac{π}{12} + \frac{1}{6}

Least Common Multiplier of

12, 6 : 12

12, 6

Least Common Multiplier (LCM)

Prime factorization of

12 : 2 \cdot 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

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

12

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{1}{6} :

multiply the denominator and numerator by

2

\frac{1}{6} = \frac{1 \cdot 2}{6 \cdot 2} = \frac{2}{12}

= - \frac{π}{12} + \frac{2}{12}

Since the denominators are equal, combine the fractions:

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

= \frac{- π + 2}{12}

x \geq \frac{- π + 2}{12} + \frac{2 π}{3} n

x \geq \frac{- π + 2}{12} + \frac{2 π}{3} n

3 x - \frac{1}{2} \leq arccos (\frac{2}{2}) + 2 πn : x \leq \frac{π + 2}{12} + \frac{2 π}{3} n

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

Simplify

arccos (\frac{2}{2}) + 2 πn : \frac{π}{4} + 2 πn

arccos (\frac{2}{2}) + 2 πn

Use the following trivial identity:

arccos (\frac{2}{2}) = \frac{π}{4}

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}{4} + 2 πn

3 x - \frac{1}{2} \leq \frac{π}{4} + 2 πn

Move

\frac{1}{2}

to the right side

3 x - \frac{1}{2} \leq \frac{π}{4} + 2 πn

Add

\frac{1}{2}

to both sides

3 x - \frac{1}{2} + \frac{1}{2} \leq \frac{π}{4} + 2 πn + \frac{1}{2}

Simplify

3 x - \frac{1}{2} + \frac{1}{2} \leq \frac{π}{4} + 2 πn + \frac{1}{2}

Simplify

3 x - \frac{1}{2} + \frac{1}{2} : 3 x

3 x - \frac{1}{2} + \frac{1}{2}

Add similar elements:

- \frac{1}{2} + \frac{1}{2} \leq 0

= 3 x

Simplify

\frac{π}{4} + 2 πn + \frac{1}{2} : 2 πn + \frac{π}{4} + \frac{1}{2}

\frac{π}{4} + 2 πn + \frac{1}{2}

Group like terms

= 2 πn + \frac{π}{4} + \frac{1}{2}

Could not simplify further

= 2 πn + \frac{π}{4} + \frac{1}{2}

3 x \leq 2 πn + \frac{π}{4} + \frac{1}{2}

3 x \leq 2 πn + \frac{π}{4} + \frac{1}{2}

3 x \leq 2 πn + \frac{π}{4} + \frac{1}{2}

Divide both sides by

3

3 x \leq 2 πn + \frac{π}{4} + \frac{1}{2}

Divide both sides by

3

\frac{3 x}{3} \leq \frac{2 πn}{3} + \frac{\frac{π}{4}}{3} + \frac{\frac{1}{2}}{3}

Simplify

\frac{3 x}{3} \leq \frac{2 πn}{3} + \frac{\frac{π}{4}}{3} + \frac{\frac{1}{2}}{3}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{2 πn}{3} + \frac{\frac{π}{4}}{3} + \frac{\frac{1}{2}}{3} : \frac{2 πn}{3} + \frac{π}{12} + \frac{1}{6}

\frac{2 πn}{3} + \frac{\frac{π}{4}}{3} + \frac{\frac{1}{2}}{3}

\frac{\frac{π}{4}}{3} = \frac{π}{12}

\frac{\frac{π}{4}}{3}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 3}

Multiply the numbers:

4 \cdot 3 = 12

= \frac{π}{12}

\frac{\frac{1}{2}}{3} = \frac{1}{6}

\frac{\frac{1}{2}}{3}

Apply the fraction rule:

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

= \frac{1}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{1}{6}

= \frac{2 πn}{3} + \frac{π}{12} + \frac{1}{6}

x \leq \frac{2 πn}{3} + \frac{π}{12} + \frac{1}{6}

x \leq \frac{2 πn}{3} + \frac{π}{12} + \frac{1}{6}

Simplify

\frac{π}{12} + \frac{1}{6} : \frac{π + 2}{12}

\frac{π}{12} + \frac{1}{6}

Least Common Multiplier of

12, 6 : 12

12, 6

Least Common Multiplier (LCM)

Prime factorization of

12 : 2 \cdot 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

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

12

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{1}{6} :

multiply the denominator and numerator by

2

\frac{1}{6} = \frac{1 \cdot 2}{6 \cdot 2} = \frac{2}{12}

= \frac{π}{12} + \frac{2}{12}

Since the denominators are equal, combine the fractions:

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

= \frac{π + 2}{12}

x \leq \frac{π + 2}{12} + \frac{2 π}{3} n

x \leq \frac{π + 2}{12} + \frac{2 π}{3} n

Combine the intervals

x \geq \frac{- π + 2}{12} + \frac{2 π}{3} n and x \leq \frac{π + 2}{12} + \frac{2 π}{3} n

Merge Overlapping Intervals

\frac{- π + 2}{12} + \frac{2 π}{3} n \leq x \leq \frac{π + 2}{12} + \frac{2 π}{3} n

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