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Popular Trigonometry >

cos^2(3x)<= 1/4

Solution

cos^{2} (3 x) \leq \frac{1}{4}

Solution

\frac{π}{9} + \frac{2 π}{3} n \leq x \leq \frac{2 π}{9} + \frac{2 π}{3} n or \frac{4 π}{9} + \frac{2 π}{3} n \leq x \leq \frac{5 π}{9} + \frac{2 π}{3} n

Interval Notation

[\frac{π}{9} + \frac{2 π}{3} n, \frac{2 π}{9} + \frac{2 π}{3} n] \cup [\frac{4 π}{9} + \frac{2 π}{3} n, \frac{5 π}{9} + \frac{2 π}{3} n]

Decimal

0.34906 \dots + \frac{2 π}{3} n \leq x \leq 0.69813 \dots + \frac{2 π}{3} n or 1.39626 \dots + \frac{2 π}{3} n \leq x \leq 1.74532 \dots + \frac{2 π}{3} n

Solution steps

cos^{2} (3 x) \leq \frac{1}{4}

For

u^{n} \leq a

, if

n

is even then

- \frac{1}{4} \leq cos (3 x) \leq \frac{1}{4}

\frac{1}{4} = \frac{1}{2}

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

Apply rule

1 = 1

= \frac{1}{2}

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

a \leq u \leq b

then

a \leq u and u \leq b

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

- \frac{1}{2} \leq cos (3 x) : - \frac{2 π}{9} + \frac{2 π}{3} n \leq x \leq \frac{2 π}{9} + \frac{2 π}{3} n

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

Switch sides

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

For

cos (x) \geq a

, if

- 1 < a < 1

then

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

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

a \leq u \leq b

then

a \leq u and u \leq b

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

- arccos (- \frac{1}{2}) + 2 πn \leq 3 x : x \geq - \frac{2 π}{9} + \frac{2 πn}{3}

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

Switch sides

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

Simplify

- arccos (- \frac{1}{2}) + 2 πn : - \frac{2 π}{3} + 2 πn

- arccos (- \frac{1}{2}) + 2 πn

Use the following trivial identity:

arccos (- \frac{1}{2}) = \frac{2 π}{3}

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{2 π}{3} + 2 πn

3 x \geq - \frac{2 π}{3} + 2 πn

Divide both sides by

3

3 x \geq - \frac{2 π}{3} + 2 πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

- \frac{\frac{2 π}{3}}{3} + \frac{2 πn}{3} : - \frac{2 π}{9} + \frac{2 πn}{3}

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

\frac{\frac{2 π}{3}}{3} = \frac{2 π}{9}

\frac{\frac{2 π}{3}}{3}

Apply the fraction rule:

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

= \frac{2 π}{3 \cdot 3}

Multiply the numbers:

3 \cdot 3 = 9

= \frac{2 π}{9}

= - \frac{2 π}{9} + \frac{2 πn}{3}

x \geq - \frac{2 π}{9} + \frac{2 πn}{3}

x \geq - \frac{2 π}{9} + \frac{2 πn}{3}

x \geq - \frac{2 π}{9} + \frac{2 πn}{3}

3 x \leq arccos (- \frac{1}{2}) + 2 πn : x \leq \frac{2 π}{9} + \frac{2 πn}{3}

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

Simplify

arccos (- \frac{1}{2}) + 2 πn : \frac{2 π}{3} + 2 πn

arccos (- \frac{1}{2}) + 2 πn

Use the following trivial identity:

arccos (- \frac{1}{2}) = \frac{2 π}{3}

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{2 π}{3} + 2 πn

3 x \leq \frac{2 π}{3} + 2 πn

Divide both sides by

3

3 x \leq \frac{2 π}{3} + 2 πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{\frac{2 π}{3}}{3} + \frac{2 πn}{3} : \frac{2 π}{9} + \frac{2 πn}{3}

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

\frac{\frac{2 π}{3}}{3} = \frac{2 π}{9}

\frac{\frac{2 π}{3}}{3}

Apply the fraction rule:

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

= \frac{2 π}{3 \cdot 3}

Multiply the numbers:

3 \cdot 3 = 9

= \frac{2 π}{9}

= \frac{2 π}{9} + \frac{2 πn}{3}

x \leq \frac{2 π}{9} + \frac{2 πn}{3}

x \leq \frac{2 π}{9} + \frac{2 πn}{3}

x \leq \frac{2 π}{9} + \frac{2 πn}{3}

Combine the intervals

x \geq - \frac{2 π}{9} + \frac{2 πn}{3} and x \leq \frac{2 π}{9} + \frac{2 πn}{3}

Merge Overlapping Intervals

- \frac{2 π}{9} + \frac{2 π}{3} n \leq x \leq \frac{2 π}{9} + \frac{2 π}{3} n

cos (3 x) \leq \frac{1}{2} : \frac{π}{9} + \frac{2 π}{3} n \leq x \leq \frac{5 π}{9} + \frac{2 π}{3} n

cos (3 x) \leq \frac{1}{2}

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

a \leq u \leq b

then

a \leq u and u \leq b

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

arccos (\frac{1}{2}) + 2 πn \leq 3 x : x \geq \frac{π}{9} + \frac{2 πn}{3}

arccos (\frac{1}{2}) + 2 πn \leq 3 x

Switch sides

3 x \geq arccos (\frac{1}{2}) + 2 πn

Simplify

arccos (\frac{1}{2}) + 2 πn : \frac{π}{3} + 2 πn

arccos (\frac{1}{2}) + 2 πn

Use the following trivial identity:

arccos (\frac{1}{2}) = \frac{π}{3}

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}{3} + 2 πn

3 x \geq \frac{π}{3} + 2 πn

Divide both sides by

3

3 x \geq \frac{π}{3} + 2 πn

Divide both sides by

3

\frac{3 x}{3} \geq \frac{\frac{π}{3}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 x}{3} \geq \frac{\frac{π}{3}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{\frac{π}{3}}{3} + \frac{2 πn}{3} : \frac{π}{9} + \frac{2 πn}{3}

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

\frac{\frac{π}{3}}{3} = \frac{π}{9}

\frac{\frac{π}{3}}{3}

Apply the fraction rule:

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

= \frac{π}{3 \cdot 3}

Multiply the numbers:

3 \cdot 3 = 9

= \frac{π}{9}

= \frac{π}{9} + \frac{2 πn}{3}

x \geq \frac{π}{9} + \frac{2 πn}{3}

x \geq \frac{π}{9} + \frac{2 πn}{3}

x \geq \frac{π}{9} + \frac{2 πn}{3}

3 x \leq 2 π - arccos (\frac{1}{2}) + 2 πn : x \leq \frac{5 π}{9} + \frac{2 π}{3} n

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

Simplify

2 π - arccos (\frac{1}{2}) + 2 πn : 2 π - \frac{π}{3} + 2 πn

2 π - arccos (\frac{1}{2}) + 2 πn

Use the following trivial identity:

arccos (\frac{1}{2}) = \frac{π}{3}

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}

= 2 π - \frac{π}{3} + 2 πn

3 x \leq 2 π - \frac{π}{3} + 2 πn

Divide both sides by

3

3 x \leq 2 π - \frac{π}{3} + 2 πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{2 π}{3} - \frac{\frac{π}{3}}{3} + \frac{2 πn}{3} : \frac{2 π}{3} - \frac{π}{9} + \frac{2 πn}{3}

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

\frac{\frac{π}{3}}{3} = \frac{π}{9}

\frac{\frac{π}{3}}{3}

Apply the fraction rule:

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

= \frac{π}{3 \cdot 3}

Multiply the numbers:

3 \cdot 3 = 9

= \frac{π}{9}

= \frac{2 π}{3} - \frac{π}{9} + \frac{2 πn}{3}

x \leq \frac{2 π}{3} - \frac{π}{9} + \frac{2 πn}{3}

x \leq \frac{2 π}{3} - \frac{π}{9} + \frac{2 πn}{3}

Simplify

\frac{2 π}{3} - \frac{π}{9} : \frac{5 π}{9}

\frac{2 π}{3} - \frac{π}{9}

Least Common Multiplier of

3, 9 : 9

3, 9

Least Common Multiplier (LCM)

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Prime factorization of

9 : 3 \cdot 3

9

9

divides by

39 = 3 \cdot 3

= 3 \cdot 3

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

3

9

= 3 \cdot 3

Multiply the numbers:

3 \cdot 3 = 9

= 9

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

9

For

\frac{2 π}{3} :

multiply the denominator and numerator by

3

\frac{2 π}{3} = \frac{2 π 3}{3 \cdot 3} = \frac{6 π}{9}

= \frac{6 π}{9} - \frac{π}{9}

Since the denominators are equal, combine the fractions:

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

= \frac{6 π - π}{9}

Add similar elements:

6 π - π = 5 π

= \frac{5 π}{9}

x \leq \frac{5 π}{9} + \frac{2 π}{3} n

x \leq \frac{5 π}{9} + \frac{2 π}{3} n

Combine the intervals

x \geq \frac{π}{9} + \frac{2 πn}{3} and x \leq \frac{5 π}{9} + \frac{2 π}{3} n

Merge Overlapping Intervals

\frac{π}{9} + \frac{2 π}{3} n \leq x \leq \frac{5 π}{9} + \frac{2 π}{3} n

Combine the intervals

- \frac{2 π}{9} + \frac{2 π}{3} n \leq x \leq \frac{2 π}{9} + \frac{2 π}{3} n and \frac{π}{9} + \frac{2 π}{3} n \leq x \leq \frac{5 π}{9} + \frac{2 π}{3} n

Merge Overlapping Intervals

\frac{π}{9} + \frac{2 π}{3} n \leq x \leq \frac{2 π}{9} + \frac{2 π}{3} n or \frac{4 π}{9} + \frac{2 π}{3} n \leq x \leq \frac{5 π}{9} + \frac{2 π}{3} n

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