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Popular >

6cos(2x-60)<= 0

Solution

6 cos (2 x - 60) \leq 0

Solution

\frac{120 + π}{4} + πn \leq x \leq \frac{3 π + 120}{4} + πn

Interval Notation

[\frac{120 + π}{4} + πn, \frac{3 π + 120}{4} + πn]

Decimal Notation

30.78539 \dots + πn \leq x \leq 32.35619 \dots + πn

Solution steps

6 cos (2 x - 60) \leq 0

Divide both sides by

6

6 cos (2 x - 60) \leq 0

Divide both sides by

6

\frac{6 cos ( 2 x - 60 )}{6} \leq \frac{0}{6}

Simplify

cos (2 x - 60) \leq 0

cos (2 x - 60) \leq 0

For

cos (x) \leq a

, if

- 1 < a < 1

then

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

arccos (0) + 2 πn \leq (2 x - 60) \leq 2 π - arccos (0) + 2 πn

a \leq u \leq b

then

a \leq u and u \leq b

arccos (0) + 2 πn \leq 2 x - 60 and 2 x - 60 \leq 2 π - arccos (0) + 2 πn

arccos (0) + 2 πn \leq 2 x - 60 : x \geq \frac{120 + π}{4} + πn

arccos (0) + 2 πn \leq 2 x - 60

Switch sides

2 x - 60 \geq arccos (0) + 2 πn

Simplify

arccos (0) + 2 πn : \frac{π}{2} + 2 πn

arccos (0) + 2 πn

Use the following trivial identity:

arccos (0) = \frac{π}{2}

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}{2} + 2 πn

2 x - 60 \geq \frac{π}{2} + 2 πn

Move

60

to the right side

2 x - 60 \geq \frac{π}{2} + 2 πn

Add

60

to both sides

2 x - 60 + 60 \geq \frac{π}{2} + 2 πn + 60

Simplify

2 x \geq \frac{π}{2} + 2 πn + 60

2 x \geq \frac{π}{2} + 2 πn + 60

Divide both sides by

2

2 x \geq \frac{π}{2} + 2 πn + 60

Divide both sides by

2

\frac{2 x}{2} \geq \frac{\frac{π}{2}}{2} + \frac{2 πn}{2} + \frac{60}{2}

Simplify

\frac{2 x}{2} \geq \frac{\frac{π}{2}}{2} + \frac{2 πn}{2} + \frac{60}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} + \frac{60}{2} : 30 + πn + \frac{π}{4}

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} + \frac{60}{2}

Group like terms

= \frac{60}{2} + \frac{2 πn}{2} + \frac{\frac{π}{2}}{2}

\frac{60}{2} = 30

\frac{60}{2}

Divide the numbers:

\frac{60}{2} = 30

= 30

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

= 30 + πn + \frac{π}{4}

x \geq 30 + πn + \frac{π}{4}

x \geq 30 + πn + \frac{π}{4}

Simplify

30 + \frac{π}{4} : \frac{120 + π}{4}

30 + \frac{π}{4}

Convert element to fraction:

30 = \frac{30 \cdot 4}{4}

= \frac{30 \cdot 4}{4} + \frac{π}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{30 \cdot 4 + π}{4}

Multiply the numbers:

30 \cdot 4 = 120

= \frac{120 + π}{4}

x \geq \frac{120 + π}{4} + πn

x \geq \frac{120 + π}{4} + πn

2 x - 60 \leq 2 π - arccos (0) + 2 πn : x \leq \frac{3 π + 120}{4} + πn

2 x - 60 \leq 2 π - arccos (0) + 2 πn

Simplify

2 π - arccos (0) + 2 πn : 2 π - \frac{π}{2} + 2 πn

2 π - arccos (0) + 2 πn

Use the following trivial identity:

arccos (0) = \frac{π}{2}

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}{2} + 2 πn

2 x - 60 \leq 2 π - \frac{π}{2} + 2 πn

Move

60

to the right side

2 x - 60 \leq 2 π - \frac{π}{2} + 2 πn

Add

60

to both sides

2 x - 60 + 60 \leq 2 π - \frac{π}{2} + 2 πn + 60

Simplify

2 x \leq 2 π - \frac{π}{2} + 2 πn + 60

2 x \leq 2 π - \frac{π}{2} + 2 πn + 60

Divide both sides by

2

2 x \leq 2 π - \frac{π}{2} + 2 πn + 60

Divide both sides by

2

\frac{2 x}{2} \leq \frac{2 π}{2} - \frac{\frac{π}{2}}{2} + \frac{2 πn}{2} + \frac{60}{2}

Simplify

\frac{2 x}{2} \leq \frac{2 π}{2} - \frac{\frac{π}{2}}{2} + \frac{2 πn}{2} + \frac{60}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{2 π}{2} - \frac{\frac{π}{2}}{2} + \frac{2 πn}{2} + \frac{60}{2} : π + 30 + πn - \frac{π}{4}

\frac{2 π}{2} - \frac{\frac{π}{2}}{2} + \frac{2 πn}{2} + \frac{60}{2}

Group like terms

= \frac{2 π}{2} + \frac{60}{2} + \frac{2 πn}{2} - \frac{\frac{π}{2}}{2}

\frac{2 π}{2} = π

\frac{2 π}{2}

Divide the numbers:

\frac{2}{2} = 1

= π

\frac{60}{2} = 30

\frac{60}{2}

Divide the numbers:

\frac{60}{2} = 30

= 30

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

= π + 30 + πn - \frac{π}{4}

x \leq π + 30 + πn - \frac{π}{4}

x \leq π + 30 + πn - \frac{π}{4}

Simplify

π + 30 - \frac{π}{4} : \frac{3 π + 120}{4}

π + 30 - \frac{π}{4}

Convert element to fraction:

π = \frac{π 4}{4}, 30 = \frac{30 \cdot 4}{4}

= \frac{π 4}{4} + \frac{30 \cdot 4}{4} - \frac{π}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{π 4 + 30 \cdot 4 - π}{4}

π 4 + 30 \cdot 4 - π = 3 π + 120

π 4 + 30 \cdot 4 - π

Group like terms

= 4 π - π + 30 \cdot 4

Add similar elements:

4 π - π = 3 π

= 3 π + 30 \cdot 4

Multiply the numbers:

30 \cdot 4 = 120

= 3 π + 120

= \frac{3 π + 120}{4}

x \leq \frac{3 π + 120}{4} + πn

x \leq \frac{3 π + 120}{4} + πn

Combine the intervals

x \geq \frac{120 + π}{4} + πn and x \leq \frac{3 π + 120}{4} + πn

Merge Overlapping Intervals

\frac{120 + π}{4} + πn \leq x \leq \frac{3 π + 120}{4} + πn

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